mm-1010 === Subject: ANN: An XML format for block designs A standard format for block designs and their properties (A proposal and an invitation for public debate) At DesignTheory.org we are developing a web-based Design Theory Resource Server for combinatorial and statistical design theory. These resources will include an online database of designs, an Encyclopaedia of Design Theory, and software packages for the generation and analysis of designs. We hope to address the needs of both researchers and practitioners of design theory. One critical element is our XML format to represent designs and their properties in a standard platform-independent manner. This will allow for the straightforward exchange of designs and their properties between various computer systems, including databases and web servers, and combinatorial, group theoretical and statistical packages. The XML format will also be used for outside submissions to our design database and to store designs in perpetuity. Our initial development is in the area of block designs, and we invite you to read and comment on our proposal for the External Representation of Block Designs, available online at: http://designtheory.org/ Please send your comments (and follow-ups) exclusively to: developers@designtheory.org This is a mailing list to which you are welcome, although not required, to join. Alternatively, you can follow the discussions through the public archives of the list. For further details, please visit: http://designtheory.org/mailing.html We will finalize the XML format for block designs after sufficient public debate, after which we shall release GAP [1], R [2], and Python [3] software for block designs. We are committed to the open source model and all products of our development will be released to the public free of charge. We shall also start developing a database of block designs, and look forward to your contributions (in the XML format) to this database. Please feel free to forward this announcement to anyone you think may be interested. [1] GAP - Groups, Algorithms and Programming http://www.gap-system.org/ [2] The R Project for Statistical Computing http://www.r-project.org/ [3] The Python programming language http://www.python.org/ === Subject: Re: High-dimensional Fredholm problems Epigone-thread: reurunvy >Hi. >Can anybody point me to examples of linear Fredholm equations of the >second kind that occur over domains of very high (even better, of >arbitrary) dimension? The examples with which I'm familiar occur >over intervals, curves, and boundaries of three-dimensional regions. I'm quite sure that the following textbook might be worthwhile to you to find what you are looking for: Sobolev, S.L. Nekotorye primeneniya funktsional'nogo analiza v matematicheskoj fizike. 'Some applications of functional analysis to mathematical physics' 3rd ed., rev. and compl., ed. by O. A. Olejnik. Moskva: Nauka. 334 p. (1988). You may order the english translation of this book at http://www.amazon.com If you do not want to buy it, I'm sure you'll find a translated copy in some university library that is within reach of a sufficiently small vicinity of the place you reside. Hope this helps M. Doerfner === Subject: Is this a famous problem? Here's a puzzle which made the rounds in 1967. Is it still well known? To me it seems more elegant and interesting than the Goldbach or Collatz Conjecture, To state the problem, one definition is needed: Y(S) = max (n / gcd(n,m)) n,m in S Problem: Find every finite subset S of the positive integers such that Y(S) = |S| or Y(S) < |S|. (|S| is the size of S.) Demonstrate that no satisfactory S has been overlooked. James Dow A (at yahoo) (This, run all together, is my e-mail address. === Subject: Number of cyclic digraphs without sources nor sinks Epigone-thread: clangshedi > A cyclic digraph is a digraph in which every vertex is in a cycle. In > other words, cyclic digraps are non-acyclic digraphs. > In a digraph, a source is a vertex with in-set of size zero; a sink is > a vertex with out-set of size zero. > I would like to know the number of labeled cyclic digraphs on n > vertices without sources nor sinks. Here is a way to count labelled digraphs in which every node belongs to a directed cycle (clearly they have neither sources nor sinks). Let K be a class of strongly connected digraphs (closed wrt relabelling the nodes) and s_n(K) denote the number of digraphs with n labelled nodes in it. a_n(K) denotes the number of all digraphs with n nodes whose strong components belong to K. Introduce the following formal generating functions: s(z,K):= sum_{n>0}s_n(K)z^n/n!, a(z,K):= sum_{n>0}a_n(K)z^n/(n!2^{n(n-1)/2}), u(z,K):= sum_{n>0}u_n(K)z^n/n! := 1-1/(1+a(z,K)) and v(z,K):= sum_{n>0}v_n(K)z^n/n!, where v_n(K):= 2^{n(n-1)/2}u_n(K). Then the following general equation takes place: s(z,K) = -log(1-v(z,K)) (*) (see V.A.Liskovets, On a general enumerative scheme for labelled graphs, Dokl. AN BSSR, 21:6 (1977), 496-499 (in Russian); MR58#21797). In particular for the class K = D of ALL strongly connected digraphs (without loops and multiple edges), _including_ the 1-node digraph [.], we have s(z,D) = -log(1-v(z,D)) and a_n(D) = 2^{n(n-1)} = #(all digraphs). Thus, a(z,D) = sum_{n>0}2^{n(n-1)/2}z^n/n!, u(z,D) = sum_{n>0}u(n)z^n/n! = 1-1/(1+a(z,D)) and v(z,D) = sum_{n>0}2^{n(n-1)/2}u_n(D)z^n/n!. s_n(D) for n=1,2,3,... is the OEIS sequence A003030 =_1_,1,18,1606,... The digraph [.] is the ONLY type of strong components whose nodes do not belong to any cycle. In order to exclude such nodes, let us eliminate [.] from D and denote C:= D-{[.]}. Then we have s(z,C) = s(z,D)-z. By (*), s(z,C) = -log(1-v(z,C)), whence v(z,C) = 1-exp(z-s(z,D)) = 1-exp(z+log(1-v(z,D))) = 1-exp(z)(1-v(z,D)). Now knowing v(z,C), we obtain u(z,C) = sum_{n>0}v_n(C)z^n/(n!2^{n(n-1)/2}) and a(z,C) = 1/(1-u(z,C))-1 = sum_{n>0}a_n(C)z^n/(n!2^{n(n-1)/2}). The coefficients a_n(C) are the desired numbers of cyclic digraphs: the ones in which all n labelled nodes belong to (directed) cycles. Numerically a_n(C) for n=1,2,3,... are 0,1,18,1699,587940, 750744901,... The difference a_4(C)-s_4(D) = 93 is verifiable easily by hand, as well as A086193(5)-a_5(C) = 5*6*9*16 = 4320 (it's evident that a_4(C) = A086193(4)). Btw, the acyclic digraphs can be counted in the same way with 1-node components only: K:= {[.]}. Besides (curiously), u_n(D) is the number of strong tournaments (A054946). Valery Liskovets === Subject: a question on probability distributions and generating functions Epigone-thread: frumgomwerl Hey, Let X and Y be two discrete nonnegative random variables. We say that X>Y, if Pr(X > t) >= Pr(Y > t), for all t>=0 Let z>0 be an arbitrary constant. Clearly, we get also : Pr(z^X > z^Y) >= Pr(z^Y > z^t) for all t, therefore z^X > z^Y. Since this is true for all z, we get that E(z^X) >= E(z^Y) for all z<1, so the generating function of X is greater than the generating function of Y for 0= E z^y for 0<=z<=1, can we conclude that X > Y? Thanx for your time, Dan === Subject: Re: a question on probability distributions and generating functions >Hey, >Let X and Y be two discrete nonnegative random variables. >We say that X>Y, if Pr(X > t) >= Pr(Y > t), for all t>=0 >Let z>0 be an arbitrary constant. Clearly, we get also : Pr(z^X > z^Y) >>= Pr(z^Y > z^t) for all t, therefore z^X > z^Y. Since this is true ??? Note that t -> z^t is _decreasing_ if 0 < z < 1 but increasing if z > 1. I think what you mean is Pr(z^X < z^t) >= Pr(z^Y < z^t) for 0 < z < 1 so z^Y > z^X >for all z, we get that E(z^X) >= E(z^Y) for all z<1, so the generating >function of X is greater than the generating function >of Y for 0= E[z^X] for 0 < z < 1. The generating function of Y is greater than that of X for 0 < z < 1. >My question is, is the other way around also true ? That is, >if E z^X >= E z^y for 0<=z<=1, can we conclude that X > Y? No. For example, let X = 1 with probability 1, while Y = 0 with probability 1/2 and 2 with probability 1/2. Since Pr(X<1) < Pr(Y<1) while Pr(X<2) > Pr(Y<2), neither XY --> E(z^X) <= E(z^Y) for z < 1, and E(z^X) >= E(z^Y) for z > 1. (Clearly at 1 we have an equality : E(1^X) = E(1^Y) = 1). So is the other way around true? I mean , i would like to know if it is true that : ---| E(z^X) <= E(z^Y) for z < 1, and | E(z^X) >= E(z^Y) for z > 1. |---> X>Y ---| (As was mentioned in your counter example we had E(z^X) <= E(z^Y) for all z, so it is not a counterexample to the rephraised quesion). Thanx, Dan >Hey, >Let X and Y be two discrete nonnegative random variables. >We say that X>Y, if Pr(X > t) >= Pr(Y > t), for all t>=0 >Let z>0 be an arbitrary constant. Clearly, we get also : Pr(z^X > z^Y) >>= Pr(z^Y > z^t) for all t, therefore z^X > z^Y. Since this is true > ??? Note that t -> z^t is _decreasing_ if 0 < z < 1 but increasing if > z > 1. I think what you mean is > Pr(z^X < z^t) >= Pr(z^Y < z^t) for 0 < z < 1 so z^Y > z^X >for all z, we get that E(z^X) >= E(z^Y) for all z<1, so the generating >function of X is greater than the generating function >of Y for 0 ... and E[z^Y] >= E[z^X] for 0 < z < 1. The generating function of > Y is greater than that of X for 0 < z < 1. >My question is, is the other way around also true ? That is, >if E z^X >= E z^y for 0<=z<=1, can we conclude that X > Y? > No. For example, let X = 1 with probability 1, while Y = 0 with > probability 1/2 and 2 with probability 1/2. Since Pr(X<1) < Pr(Y<1) > while Pr(X<2) > Pr(Y<2), neither X generating functions E[z^X] = z and E[z^Y] = (1+z^2)/2 satisfy > E[z^X] <= E[z^Y] for all z. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Rephraising of the question >We know that X>Y --> E(z^X) <= E(z^Y) for z < 1, and > E(z^X) >= E(z^Y) for z > 1. >(Clearly at 1 we have an equality : E(1^X) = E(1^Y) = 1). >So is the other way around true? I mean , i would like to know if it >is true that : > ---| >E(z^X) <= E(z^Y) for z < 1, and | >E(z^X) >= E(z^Y) for z > 1. |---> X>Y > ---| No. Consider the case where X = 1 with probability 3/4, 3 with probability 1/4 Y = 0 with probability 1/4, 2 with probability 3/4 E[z^X]-E[z^Y] = -1/4 + 3/4 z - 3/4 z^2 + 1/4 z^3 = -(1-z)^3/4 < 0 for z < 1 and > 0 for z > 1. But Pr(X<2) > Pr(Y<2), so it's not true that X>Y Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Polynomials times a Gaussian are dense in L^2 > What's an easy low-brow way to prove that polynomial > functions times the Gaussian exp(-x^2) are dense in L^2(R)? The way we were given in lectures was as follows. Since the trigonometric polynomials are uniformly dense in C[a,b] it is easy to see that the space of finite linear combinations of functions of the form: e^(iax)e^(-x^2/2) (a = 0,1,2,...) is uniformly dense in C_0(R) (the space of continuous functions on R functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is uniformly dense in C_0(R). Now given f in L^2(R) first approximate it in L^2 by continuous g with compact support (as in previous messages) then uniformly approximate g(x)e^(x^2/2) by functions of the form p(x)e^(-x^2/2) where p is a polynomial. It is easy to see this gives an L^2 approximation of g by functions of the form p(x)e^(-x^2) so we're done. Michael === Subject: Re: Polynomials times a Gaussian are dense in L^2 >> What's an easy low-brow way to prove that polynomial >> functions times the Gaussian exp(-x^2) are dense in L^2(R)? >The way we were given in lectures was as follows. Since the >trigonometric polynomials are uniformly dense in C[a,b] it is easy to >see that the space of finite linear combinations of functions of the >form: >e^(iax)e^(-x^2/2) (a = 0,1,2,...) >is uniformly dense in C_0(R) (the space of continuous functions on R >tending to 0). That's clear. >functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is >uniformly dense in C_0(R). How does one deduce that, exactly? >Now given f in L^2(R) first approximate it in L^2 by continuous g with >compact support (as in previous messages) then uniformly approximate >g(x)e^(x^2/2) by functions of the form p(x)e^(-x^2/2) where p is a >polynomial. It is easy to see this gives an L^2 approximation of g by >functions of the form p(x)e^(-x^2) so we're done. >Michael ************************ David C. Ullrich === Subject: Re: Polynomials times a Gaussian are dense in L^2 >>The way we were given in lectures was as follows. Since the >>trigonometric polynomials are uniformly dense in C[a,b] it is easy to >>see that the space of finite linear combinations of functions of the >>form: >>e^(iax)e^(-x^2/2) (a = 0,1,2,...) Oops, meant to say a is an integer rather than non-negative integer. In fact the rest of the proof goes through fine if you only restrict a to being in R. >>is uniformly dense in C_0(R) (the space of continuous functions on R >>tending to 0). >That's clear. >>functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is >>uniformly dense in C_0(R). >How does one deduce that, exactly? Well it suffices to prove e^(iax)e^(-x^2/2) can be uniformly approximated on R by p(x)e^(-x^2/2). Let p_n be the first n terms of the Taylor series of e^(iax). By the triangle inequality: |e^(iax)e^(-x^2/2)-p_n(x)e^(-x^2/2)| <= e^(|a||x| - x^2/2) so this is < epsilon outside some compact interval I, for all n. But p_n(x) -> e^(iax) uniformly on I so we're done. Michael === Subject: Re: Polynomials times a Gaussian are dense in L^2 >The way we were given in lectures was as follows. Since the >trigonometric polynomials are uniformly dense in C[a,b] it is easy >see that the space of finite linear combinations of functions of the >form: >e^(iax)e^(-x^2/2) (a = 0,1,2,...) >Oops, meant to say a is an integer rather than non-negative integer. >In fact the rest of the proof goes through fine if you only restrict a >to being in R. >is uniformly dense in C_0(R) (the space of continuous functions on R >tending to 0). >>That's clear. >functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is >uniformly dense in C_0(R). >>How does one deduce that, exactly? >Well it suffices to prove e^(iax)e^(-x^2/2) can be uniformly >approximated on R by p(x)e^(-x^2/2). Let p_n be the first n terms of >the Taylor series of e^(iax). By the triangle inequality: >|e^(iax)e^(-x^2/2)-p_n(x)e^(-x^2/2)| <= e^(|a||x| - x^2/2) Well duh. Took me a second to see why that inequality was true, but yes, it's clear. This is the simple proof I was coming to think didn't exist. Keen. >so this is < epsilon outside some compact interval I, for all n. But >p_n(x) -> e^(iax) uniformly on I so we're done. >Michael ************************ David C. Ullrich === Subject: Re: Polynomials times a Gaussian are dense in L^2 >> What's an easy low-brow way to prove that polynomial >> functions times the Gaussian exp(-x^2) are dense in L^2(R)? The problem is harder than it looks. The property of the function exp(-x^2) which is needed is that the corresponding moment problem must have a unique solution. This looks like an L_1 condition, but everything between 1 and 2 is equivalent for this problem, and the usual nasc conditions can easily be obtained by L_2 arguments. This is easy, but lengthy, so I do not think it is what is being desired. It can be done with only real analysis, but using complex numbers. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Polynomials times a Gaussian are dense in L^2 >What's an easy low-brow way to prove that polynomial >functions times the Gaussian exp(-x^2) are dense in L^2(R)? >I know a general-purpose L^2 Stone-Weierstrass theorem >that does the job, but it requires knowing the Fourier >transform is unitary, which takes a while to prove. >I also know the eigenfunctions of the harmonic oscillator >Hamiltonian are an orthonormal basis of L^2(R), and this >also does the job, but again it relies on some stuff that >takes work to prove. >For a class I plan to teach, I'd like a quick proof >that doesn't use much machinery, if one exists. It's possible that there is no proof as elementary as what you want. The following fact surprised me: Fact: There exists a function G such that (*) G > 0, G is continuous, and x^n G(x) -> 0 as x -> infinity, for n = 0, 1, ..., such that the functions P(x) G(x) are _not_ dense in L^2(R) (note that the fact that x^n G(x) -> 0 for all n shows that P(x) G(x) is in L^2 for every P.) My point is that thinking about a totally low-brow proof of what you want I assumed it would follow just from the fact that G(x) = exp(-x^2) satisfies (*). But it doesn't follow from (*), so any proof must use some more special property of exp(-x^2). (In particular it's not clear that, for example, Nemo's argument can be fixed - that was the sort of argument I was looking for for a while...) Proof of the Fact: Let phi be infinitely differentiable on R, with support(phi) contained in [1,2] (and phi not identically 0.) Let F^ = phi (where F^ is the Fourier transform of F.) Then F is a Schwarz function, so in particular F(x) (1 + |x|^n) is in L^2 for n = 0, 1, ... . So we can choose numbers c_n > 0 such that f is in L^2(R), where f(x) = F(x) sum(c_n (1 + |x|^n)) (and also such that sum(c_n (1 + |x|^n)) is continuous, in case that doesn't follow.) Let G(x) = 1 / sum(c_n (1 + |x|^n)). It's clear that G satisfies (*). But the P(x) G(x) are not dense in L^2, because f is orthogonal to all of them: = = linear combination of derivatives of F^ at the origin = 0. ************************ David C. Ullrich === Subject: Re: Polynomials times a Gaussian are dense in L^2 >> What's an easy low-brow way to prove that polynomial >> functions times the Gaussian exp(-x^2) are dense in L^2(R)? >Suppose that f is in L^2(R) and is orthogonal to all such products. >Expanding exp(zx) in a power series and using Fubini's theorem, check >that >int_R f(x) exp(zx) exp(-x^2) dx = 0, >first for all complex z of sufficiently small modulus, then for all >complex z by analytic continuation. In particular, >int_R f(x) exp(-itx) exp(-x^2) dx = 0, >for all real t. As f(x)exp(-x^2) is clearly an element of L^1(R), >this last display means that the Fourier transform of f vanishes. >Thus f = 0, almost everywhere. Yeah, I thought of suggesting more or less that (don't see why we need to restrict to z of small modulus.) But it uses the uniqueness theorem for the Fourier transform; since the OP specified he didn't want to use the Plancherel theorem I decided this was insufficiently low-brow. There must be a proof that doesn't depend on anything remotely non-trivial about the Fourier transform, along the lines of Nemo's almost-proof. Not that I see what it is... ************************ David C. Ullrich === Subject: category question Is there a term for a category whose morphisms can be expressed in terms of composion through a single object? Namely, there's an object A in the category such that for objects B and C, Hom(B,C) is Hom(B,A) composed with Hom(A,C). Ie, the morphisms to and from A generate all morphisms in the category. I'm also looking for a more general version, a category where the morphisms to and from a finite set of objects generate the morphisms in the category. Any help included you. Karl Hallowell khallow@hotmail.com === Subject: Re: category question |Is there a term for a category whose morphisms can be expressed in |terms of composion through a single object? Namely, there's an object |A in the category such that for objects B and C, Hom(B,C) is Hom(B,A) |composed with Hom(A,C). Ie, the morphisms to and from A generate all |morphisms in the category. I'm also looking for a more general |version, a category where the morphisms to and from a finite set of |objects generate the morphisms in the category. Any help included |you. when every morphism in category d factors through an object in full subcategory e, d is called a full subcategory of the karoubi envelope of e. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Goldbach-Idea - probably dumb! I remember reading in a book, I think Recreations in the Theory of Numbers by Dover, about lucky numbers, which are derived using something like the Sieve of Eratosthenes, but with a different rule to strike out the numbers. The numbers surviving the sieve are lucky and include some primes and some composites. If I remember correctly, the book said that the lucky numbers have properties similar to the primes as far as their distribution, twin lucky numbers, and also obey a Goldbach-like Conjecture. > Just thought I'd throw in my 2 cents on Goldbach. Not my area, but the > thread made me think of the following. > If you were to generate a set of false primes by insisting that in > any interval they are distributed amongst the odd numbers with the > same frequency as the real ones (we can forget about 2), > What is the probability that such a set has the Goldbach property > (i.e. any even number can be expressed as the sum of two false > primes)? Is it non-zero? What would the limit of this probability be > if one only had to start looking after an arbitrarily large n? > My really stupid idea was that if (subject to some reasonable > verification method for sets of false primes being stated) the > probability was non-zero, then the Goldbach conjecture could be true > by accident. > If however the probability does tend to zero, then surely the > conjecture is equally fascinating! > I'm no number theorist, but would love to know if anyone's thought > along these lines. === Subject: Re: Goldbach-Idea - probably dumb! > I remember reading in a book, I think Recreations in the Theory of > Numbers by Dover, about lucky numbers, which are derived using > something like the Sieve of Eratosthenes, but with a different rule to > strike out the numbers. The numbers surviving the sieve are lucky > and include some primes and some composites. If I remember correctly, > the book said that the lucky numbers have properties similar to the > primes as far as their distribution, twin lucky numbers, and also obey > a Goldbach-like Conjecture. Indeed for a whole class of sieve generated sequences one can prove a prime number theorem. See some papers (from the 70s, I think) by Wunderlich, et al. I seem to recall (and this just may be wishful thinking) that there is even a Riemann hypothesis that goes along with this that is true. Don > Just thought I'd throw in my 2 cents on Goldbach. Not my area, but the > thread made me think of the following. > > If you were to generate a set of false primes by insisting that in > any interval they are distributed amongst the odd numbers with the > same frequency as the real ones (we can forget about 2), > > What is the probability that such a set has the Goldbach property > (i.e. any even number can be expressed as the sum of two false > primes)? Is it non-zero? What would the limit of this probability be > if one only had to start looking after an arbitrarily large n? > > My really stupid idea was that if (subject to some reasonable > verification method for sets of false primes being stated) the > probability was non-zero, then the Goldbach conjecture could be true > by accident. > > If however the probability does tend to zero, then surely the > conjecture is equally fascinating! > > I'm no number theorist, but would love to know if anyone's thought > along these lines. === Subject: when gcd does not exist? Epigone-thread: roahaithang I started reading Ireland and Rosen's A Classical Introduction to Modern Number Theory and noticed a remark that in an arbitrary ring the greatest common divisor not necessarily exists. I couldn't exhibit an example. Could you give an example of integral domain such that there exist two elements a,b, but there is no their greatest common divisor? An element d in R is called the greatest common divisor (gcd) of elements a and b from R, if (a) d|a and d|b; (b) d'|a and d'|b implies d'|d. If R is integral domain, then any two gcd's of a and b are associated (differ by a unit). === Subject: Re: when gcd does not exist? [student] | Could you give an example of integral domain such that there exist two | elements a,b, but there is no their greatest common divisor? Magidin gave you one example, here is a similar one, which you may like better, depending on your taste: Let k be a field and consider the integral domain k[x,y,z,w] = k[X,Y,Z,W]/(XY-ZW). Let a = xy = zw and b = xz. Then both x and z are common divisors, they are both of maximal degree, and they are distinct. Both are maximal, but none of them is greatest. (Since the relation is homogeneous, the ring has an obvious grading which we can use for the same purpose as the norms in Magidin's example.) -- Stein Arild StrÀmme +47 55584825, +47 95801887 Universitetet i Bergen Fax: +47 55589672 Matematisk institutt www.mi.uib.no/stromme/ Johs Brunsg 12, N-5008 BERGEN stromme@mi.uib.no === Subject: Re: when gcd does not exist? >I started reading Ireland and Rosen's A Classical Introduction to >Modern Number Theory and noticed a remark that in an arbitrary ring >the greatest common divisor not necessarily exists. I couldn't exhibit >an example. >Could you give an example of integral domain such that there exist two >elements a,b, but there is no their greatest common divisor? Not that it matters, as there are examples with integral domains, but you do notice the difference between arbitrary ring and integral domain? >An element d in R is called the greatest common divisor (gcd) of >elements a and b from R, if >(a) d|a and d|b; >(b) d'|a and d'|b implies d'|d. >If R is integral domain, then any two gcd's of a and b are associated >(differ by a unit). Let R = Z[sqrt(-5)], a=6, b=2+2*sqrt(-5). The elements of R are of the form x+y*sqrt(-5) with x and y integers, and with the obvious addition and multiplication. Both a and b are multiples of both 2 and 1+sqrt(-5). So a gcd for a and b would have to be a common multiple of 2 and 1+sqrt(-5). Consider the norm function, N:Z[sqrt(-5)] -> Z, given by N( x+y*sqrt(-5) ) = x^2 + 5y^2. It is an easy exercise to verify that: (i) N is multiplicative: N(r*s) = N(r)N(s) for all r,s in Z[sqrt(-5)]. (ii) If r divides s in Z[sqrt(-5)], then N(r) divides N(s) in Z. (iii) r is a uni in Z[sqrt(-5)] if and only if N(r)=1, if and only if r=1 or r=-1. N(6) = 36 and N(2+2*sqrt(-5))=24. So a gcd would have norm dividing gcd(36,24)=12. Since N(2)=4 and N(1+sqrt(-5))=6, any common multiple of 2 and 1+sqrt(-5) has norm a multiple of 12. So if a gcd exists for a and b, then its norm must be exactly equal to 12. However, there is no integer solution to x^2+5y^2 = 12; clearly we need |y|<=1; if y=0 then we would need x^2=12, impossible, and if |y|=1 we would need x^2=7, also impossible. So a and b have no gcd in Z[sqrt(-5)]. (A bit more complicated: since x+y*sqrt(-5) is a multiple of 2, you can write (2n) + (2m)*sqrt(-5); the norm would then be 4n^2 + 20m^2, so m=0, n^2 = 3, impossible) It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: a question on probability distributions and generating functions Epigone-thread: frumgomwerl Dan asks (after corrections): Let X and Y be two discrete nonnegative random variables. We say that X>Y, if Pr(X > t) >= Pr(Y > t), for all t>=0 (*) Let z>0 be an arbitrary constant. Clearly, we get also : Pr(z^X > z^t) >= Pr(z^Y > z^t) for all t if z>1, Pr(z^X > z^t) <= Pr(z^Y > z^t) for all t if z<1. So E(z^X)>=E(z^Y) if z>1 and E(z^X)<=E(z^Y) if 0Y (look at t=1.5). Don Coppersmith === Subject: Minimising near diagonal terms in a matrix Message-id: <3F576391.2090403@cui.unige.ch> Here is a pb I'd like to submit, which I guess is close to the Metric TSP and wonder whether some of you may know a solution (exact or approx). Given an interdistance matrix (Aij=d(Xi,Xj) for some Xi's), what is the best permutation so that given r>0 the sum, for every line of the matrix, of the r terms around the diagonal is minimum? I have experimented with the greedy algo and a DP procedure for r=1 (which gives good result) but would like to extend to the case r>1. I that a well-known problem? I gess so but cannot find any good reference... Stephane -- Dr. Stephane Marchand-Maillet (MER) Head of the Viper group marchand@cui.unige.ch http://viper.unige.ch/ Computer Vision and Multimedia Lab, CUI, University of Geneva 24 Rue du General-Dufour - 1211 Geneva 4 - Switzerland Tel: +41 (0)22 705 7631 / +41 (0)22 705 7660 Fax: +41 (0)22 705 7780 http://www.mrml.net/ http://www.benchathlon.net/ === Subject: Re: Q: Norm of difference of fractional powers of operators I found a reference for this inequality: Rajendra Bhatia Perturbation Inequalities for the Absolute Value Map in Norm Ideals of Operators J. Operator Theory 19 (1988), 129-136 where the inequality is actually proven. Markus === Subject: Re: Think Small game > In the three-player case, there is a Nash equilibrium with player 1 > always choosing 1, player 2 choosing 2 and player 3 choosing 3. For that matter, it's also an equilibrium if Player 1 always chooses 1, Player 2 always chooses 2, and Player 3 always chooses 17. > This shows one of the pitfalls of the notion of Nash equilibrium in > multi-player games. Well, you can also ask whether the equilibrium is stable, which this one isn't. The stable equilibria for this game are more interesting, I think. David desJardins === Subject: Re: Qualities of a good math teacher >Hello Everyone!! Mathematics is concerned with the nature of numbers, geometric objects, and other mathematical concepts; it is concerned with their cognitive origins and with their application to reality, it deals the validation of methods of mathematical inference, it deals with the logical problems associated with mathematical infinitude. It is a model tool of logical perfection, because of the clarity of its concepts and the certainty of its conclusions. So a good teacher therefore should devote much effort to explaining the nature of mathematics to students. He should provide an introduction to the major issues in the field of mathematics, and the historically important views on mathematical issues. All citizen of an advanced country should have some familiarity with mathematics as prerequisite for thinking about mathematical way for decision making ways, and application of this great subject. And it is rewarding. A great teacher is inspired student at the beginning, and end up in a remembrance of a man of credit to the memories of his student. Irrespective of countries, clouts and time if mathematicians down the history may influence you, you may become a good mathematician and if you have your beautiful way of influencing others to think and do mathematics, then you are a good teacher. Tapas K.Bhattacharya.MIEEE. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: combinations concept > I am doing a short research paper on how to teach combinations. Here > in Malaysia, we teach permutation and combination to students in Form > 5 ( age of 17-18 ). Students always have problems with this topic. I > wonder if students in other countries facing this problem? How > teachers teach this topic effectively? Students always feel frustrated > when they fail to get the correct answers. I would like to investigate > if CAI can help by using the elaborative feedbacks to give extra > explanation. Could someone please give me some comments how to teach > this topic effectively? I greatly appreciate it. I have taught permutations and combinations as part of an Applied Discrete Math course to freshmen in college in the US (about one year older than your group). They have not had too much trouble with the concepts, though that may be because only engineering students are bothering to take the class---the math-phobic are not there. Good lectures and lots of examples seem to be the key to students learning this material. Students who are used to memorizing algorithms generally do more poorly at these problems than those who have learned problem-solving skills. Solving permutations and combinations problems often requires coming up with the appropriate representation, after which the application of the formulas is rather trivial. There are only three formulas worth memorizing for these problems: n!, n choose k, and n^k. The trick is in learning how to apply them. ------------------------------------------------------------ Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus Professor of Biomolecular Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Affiliations for identification only. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Numb3rs > Has anyone heard anything about > this new show on CBS called Numb3rs? > I've seen a couple of commercials for it > (starts in January) and it looks pretty cool > --like they use numbers and patterns to solve > crimes. Also in January, teachers, grading the New York Regents Exams, will be doing something similar. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Limit n^(1/n) = ? when n approaches +infinite >>limit n^(1/n) = ? when n approaches +infinite >>The answer is 1. >>How come? > How come indeed. It looks like 0 to me. No, it goes to 1. Try it, square root(2), cube root (3), fourth root (4) (=square root (2)), fifth root (5),... If you take the logarithm, you get (log n)/n and since log n and n are of the same order, they go to infinity at the same rate and the ratio is 1. Say we set n^(1/n) = p. As p approaches 1 from the positive direction, p^n as n approaches infinity can be arbitrarily large, so there isn't any significant difference between p^n and n and p^n/n = 1. --Jeff -- It is only those who have neither fired a shot nor heard the shrieks and groans of the wounded who cry aloud for blood, more vengeance, more desolation. War is hell. --William Tecumseh Sherman -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Limit n^(1/n) = ? when n approaches +infinite > limit n^(1/n) = ? > when n approaches +infinity > The answer is 1. > How come? lim n^(1/n) = lim e^(ln(n^(1/n))) = lim e^((1/n)ln(n)) = lim e^(ln(n)/n) = e^lim(ln(n)/n) (by continuity) = e^lim(1/n / 1) (by L'Hopital's Rule) = e^0 = 1. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Limit n^(1/n) = ? when n approaches +infinite > limit n^(1/n) = ? when n approaches +infinite > The answer is 1. > How come? > I think it for a long time. I cannot solve it. > Can someone help me or give me a hint? > I really want to know the solution. > Dan Here's a BIG hint: Find, using l'Hospital's rule, lim x -> infinity {ln(x^(1/x))}. That is, take the natural log of n^(1/n), and find THAT limit as n -> infinity. I sure HOPE that the homework do-ers don't give you the complete solution before you have a chance to try this. -- Delete the second o to email me. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Limit n^(1/n) = ? when n approaches +infinite > limit n^(1/n) = ? when n approaches +infinite > The answer is 1. > How come? > I think it for a long time. I cannot solve it. > Can someone help me or give me a hint? > I really want to know the solution. n^(1/n) = e^ (ln(n)/n) lim n->+infinity e^(ln(n)/n) = e^ lim n->+infinity ln(n)/n = e ^ 0 = 1 Note: I skipped the most important step, proving that ln(n)/n -> 0 as n -> +infinity. ------------------------------------------------------------ Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus Professor of Biomolecular Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Affiliations for identification only. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Help > Can someone please solve this problem- > 2,236,000-17,740Y=3,500,000 I can solve it, can you try? --Jeff -- It is only those who have neither fired a shot nor heard the shrieks and groans of the wounded who cry aloud for blood, more vengeance, more desolation. War is hell. --William Tecumseh Sherman -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Trigonometry - Proving trigonometric identities > How do you prove the following identity: > cosA * cotA + sinA = cscA First, cos^2(A) + sin^2(A) = 1 and sin(A)/sin(A) = 1. So equate these: cos^2(A) + sin^2(A) = sin(A)/sin(A). Divide by sin: cos(A) 1 cos(A)------ + sin(A) = ------. sin(A) sin(A) Or cos(A)cot(A) + sin(A) = csc(A). -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Trigonometry - Proving trigonometric identities > How do you prove the following identity: > cosA * cotA + sinA = cscA Put everything in terms of sinA and cosA and get a common denominator on the left side of the equation. --Jeff -- It is only those who have neither fired a shot nor heard the shrieks and groans of the wounded who cry aloud for blood, more vengeance, more desolation. War is hell. --William Tecumseh Sherman -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Trigonometry - Proving trigonometric identities > How do you prove the following identity: > cosA * cotA + sinA = cscA The left side = cosA(cosA/sinA) + sinA = (cosA)^2 / sinA + (sinA)^2/sinA = [(cosA)^2 + (sinA)^2] / sinA = 1/sinA = cscA -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Trigonometry - Proving trigonometric identities > How do you prove the following identity: > cosA * cotA + sinA = cscA Try multiplying both sides by sin(A) and simplifying the fractions---see it it looks more familiar then. ------------------------------------------------------------ Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus Professor of Biomolecular Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Affiliations for identification only. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Posting Rules? I'm a newbie here. Where can I find the posting rules for this group? LEEinSC_7007 The Moderator responds: As noted in all messages posted to this newsgroup, in the footnote appended to the bottom of all messages, the newsgroup web site is at http://www.thinkspot.net/k12math/ There you can find a link to the newsgroup charter which has the definitive rules for participating in the newsgroup k12.ed.math. Additionally, you will find a link to the Faq-O-Matic which contains additional guidelines for participation in this newsgroup. k12.ed.math moderator kem-moderator@k12groups.org http://www.thinkspot.net/k12math/ -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: terms in Math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmj09356; >Can anyone tell me the URL with pages devoted to math term's. One such is http://mathworld.wolfram.com/ === Subject: Re: coverings of a Mobius band by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmT09372; >Are the even degree covers of a Mobius band annulus(es)? Are the odd degree covers of a Mobius band, Mobius bands again? What is the universal cover of a Mobius band? Yes, yes, and R x [-1, 1]. One can imagine a strip of paper with n half-twists before gluing the ends as the total space of the degree n cover of the Mobius band. The strip of paper is I x [-1, 1], and the gluing map for taping the ends is the homeomorphism (-1)^n * - : [-1, 1] --> [-1, 1], so we get an annulus for n even and a Mobius strip for n odd. For the universal cover, imagine an infinite strip with infinitely many half-twists. Todd Trimble === Subject: Re: The Possible Cure For AIDS in part: > There are no A's, G's, 4's, O's, >>The G is gimmel the 3 rd letter of the Hebrew alphabet. >>The vowels (mplicit in old Hebrew but made explicit by the Masorites) >>include an ah sound and a oh sound. >AFAIK the Torah codes are never done with the vowels. That's quite true. Including vowels makes it harder to find matches. But if someone wanted to find English words in the Tanakh, the standard convention is, of course, Aleph for A and Ayin for O. As for the number 404, two letters would represent that in standard Hebrew numeric notation: Tau and Daleth. But Hebrew has a word for silver, although *ancient* Hebrew hardly had a word for oxygen, even if it certainly has one today. Actually, we should be asking what letters of the Hebrew alphabet are supposed to stand for and ! John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Possible Cure For AIDS > But Hebrew has a word for silver, although *ancient* Hebrew hardly had a Kessef. > word for oxygen, even if it certainly has one today. Khamtzan which means sour-er. This corresponds to the German Saurstuff. > Actually, we should be asking what letters of the Hebrew alphabet are > supposed to stand for and ! There is a typefont used to crossreference passages in the Talmud to the Shulkhan Arooch. It looks like superscripts because of the way it is printed. This was originated about 1500 of the common era. Bob Kolker === Subject: Re: The Possible Cure For AIDS in part: >There are no A's, G's, 4's, O's, etc. in Hebrew. Did you transliterate >these according to some arbitrary scheme, or did you search for Hebrew >letter strings? Not that it matters, of course. But while we're talking about a cure for AIDS... Some years back, there were news stories about how, for AIDS research, modified mice were created with human immune systems. This was done by taking cells from human fetuses. Naturally, this was controversial because of the abortion issue. I noted that nobody cares about mouse fetuses. If you can fix a mouse so that it _can_ get AIDS by giving it a human being's immune system, it would seem you could fix a human so that he can't get AIDS by giving him a mouse's immune system. There were probably very good reasons why this couldn't _really_ be done, but I had thought it worth mentioning. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Beal's conjecture > => A large prize is offered by banker Beal for a solution to > => the Beal Conjecture: the equation x^p + y^q = z^r has no solutions > => for p, q, r > 2 and coprime integers x, y, z. > = => Sorry if this has been discussed here - my only justification > => is that I just recently discovered this group but I am curious to > => know if Wile's proof of FLT also covers the Beal Conjecture > => Or have any counter-examples been found? Wile's proof covered the case p=q=r and relaxed the coprime requirement for x, y, & z. IE: There is no integer solution when p=q=r whatever the coprimality of x, y, & z. tom -- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an hour. === Subject: Mathforge.net :: Near-numbers, autonomic computing, John Derbyshire, and Church-Turing The Latest Math News from Mathforge.net http://mathforge.net ****An invitation to additive prime number theory**** A.V. Kumchev and D.I. Tolev have compiled a short document entitled An Invitation to Additive Prime Number Theory[~60pp, pdf]. The document serves as an introductory guide to graduate-level students on... http://mathforge.net/index.jsp?page=seeReplies&messageNum=981 ****Are there encoded messages in the Bible?**** researchers both supporting and denying the statistical evidence for 'hidden messages' found in the Old Testament tell their tales. Some... http://mathforge.net/index.jsp?page=seeReplies&messageNum=974 ****Near-numbers: the new 'limit'**** There's a very interesting paper by Frank J Swenton of Middlebury College called Limits and the System of Near-Numbers[19pp, pdf]. What seemed at first glance (at the title and abstract) like an uninspired... http://mathforge.net/index.jsp?page=seeReplies&messageNum=970 ****Introduction to autonomic computing**** While not itself mathematical in nature, the concept is built around ideas garnered from areas of artificial intelligence research and it is easy to see that autonomic computing could have applications... http://mathforge.net/index.jsp?page=seeReplies&messageNum=969 ****Derbyshire's Diary**** John Derbyshire, author of Prime Obsession, has a mathematical problem accompanying each of his Diary entries located in his Web Journalism folder. If you sift through enough of the partisan propaganda... http://mathforge.net/index.jsp?page=seeReplies&messageNum=968 ****Maple 9.5 Released**** MapleSoft has released version 9.5 of their popular symbolic and numeric computational software suite Maple. New Features include added packages (optimization, logic, and root finding), OpenMaple access... http://mathforge.net/index.jsp?page=seeReplies&messageNum=967 ****Mathematica 5.1 released**** The Mathematica version has jumped a tenth of a point, and Wolfram has added scores of new features to the new release, including Web Services support, a benchmarking package, string manipulation functions,... http://mathforge.net/index.jsp?page=seeReplies&messageNum=966 ****Quantum computers and the Church-Turing Thesis**** The original Church-Turing Thesis states that every function which would naturally be regarded as computable can be computed by a Turing Machine and Petrus H. Potgeiter mentions in his paper Zeno Machines... http://mathforge.net/index.jsp?page=seeReplies&messageNum=963 ****Sobering U.S. Student Math Scores**** In a disheartening follow-up to the Putnam story below, news (Seattle Times) outlets (New York Times)everywhere are reporting the horrid state of U.S. student math skills. The Organisation for Economic... http://mathforge.net/index.jsp?page=seeReplies&messageNum=959 ''Mathforge ran a story about the Putnam Competition... http://mathforge.net/index.jsp?page=seeReplies&messageNum=958 === Subject: What kind of matrix can map positive element vectors to positive element vectors? Hi all, Suppose I have a matrix A, and a positive element vector x, then y=A*x, I want y to be also positive element vector... What can I say about A? I want all such kinds of A? What kinds of A can let me have both directions: x positive elements <=> y=A*x positive elements? === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > Suppose I have a matrix A, and a positive element vector x, > then y=A*x, > I want y to be also positive element vector... > What can I say about A? I want all such kinds of A? > What kinds of A can let me have both directions: > x positive elements <=> y=A*x positive elements? Take a simple case and try to learn from it. The unit vectors for x will pull out the columns of A. Sure seems to say that A must at least have all of its elements positive. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > > Suppose I have a matrix A, and a positive element vector x, > > then y=A*x, > > I want y to be also positive element vector... > > What can I say about A? I want all such kinds of A? > > What kinds of A can let me have both directions: > > x positive elements <=> y=A*x positive elements? > > > > Take a simple case and try to learn from it. > The unit vectors for x will pull out the columns of A. > Sure seems to say that A must at least have all of its > elements positive. Necessary and sufficient. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? >> Hi all, >> Suppose I have a matrix A, and a positive element vector x, >> then y=A*x, >> I want y to be also positive element vector... >> What can I say about A? I want all such kinds of A? >> What kinds of A can let me have both directions: >> x positive elements <=> y=A*x positive elements? >> Take a simple case and try to learn from it. >> The unit vectors for x will pull out the columns of A. >> Sure seems to say that A must at least have all of its >> elements positive. >Necessary and sufficient. Neither necessary (unless your positive means >= 0) nor sufficient. Note that lucy wanted <=>. I'll assume A is a square matrix. It's easy to see A must be nonsingular, else given vector x > 0 (i.e. all x_i > 0) with Ax > 0, you could add to x a suitable multiple of a vector w with Aw = 0 to get a vector v not > 0 with Av = Ax > 0. A must map the nonnegative cone C = {x in R^n: all x_i >= 0} onto itself. Note that the extreme rays of C pass through the standard unit vectors e(j) (with e(j)_i = 1 if i=j, 0 otherwise). That is, these are the only members w of C such that if w = t x + (1-t) y with x,y in C and 0 < t < 1, then x and y are scalar multiples of w. Now A must map extreme rays to extreme rays, and from this it's easy to see that A must be of the form A = D P where P is a permutation matrix and D a diagonal matrix with positive elements on the diagonal. Conversely, everything of this form has the desired property. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: How to determine parameter integer values such that quadratic has integer solutions? Hi all, In particular, I'm interested in finding the positive integer values of g such that the following quadratic in v has integer roots: v^2 - 7v + (12 - 12g) Evidently this requires that 1 + 48g be a perfect square. I can have Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, 20, 35, ... This sequence is more conveniently summarized as something-mod-something-else, but I forget exactly what the two somethings are. How to figure this out? cdj === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) v = (7 +- sqr(49 - 48 + 48g)/2 = (7 +- sqr(1 + 48g)/2 sqr(1 + 48g) must be odd number, 2n + 1 1 + 48g = 4n^2 + 4n + 1 12g = n^2 + n = n(n + 1) Case 12 | n. Done Case 4 | n, not 3 | n. 3 | n+1; n = 3k - 1 Case 3 | n, not 2 | n. 4 | n+1; n = 4k - 1 Case 2 | n, not 4 | n. Not possible Three family parametrization of solutions. n = 12k n = 3k - 1 provided 4 | n n = 4k - 1 provided 3 | n Expect some duplicates. > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? The square root of 1+48g is odd... for fuck's sake I'm blind.... === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? cdj: Here's an outline: We know that 1+48g is a perfect square. Since g is an integer, 48g is even, and 1+48g is odd. So, the square root of 1+48g is odd, and we can write that square root as 2n+1 for some integer n. Thus, (2n+1)^2 = 1+48g Expanding and simplifying, 4n^2+4n+1 = 1+48g 4n^2+4n = 48g n^2+n = 12g We need a little number theory here: Consider the above equation (mod 12): (n^2+n) (mod 12) = 0 Trying the 11 possibilities, we find that n (mod 12) must be 0, 3, 8, or 11, that is, we must be able to write n as 12k or 12k+3 or 12k+8 or 12k+11 for some integer k. (The above condition is equivalent to (n^2+n) = 0 (mod 3) AND (n^2+n) = 0 (mod 4), so with some work, we really need only to consider 3+4 = 7 cases.) Substituting for the first possibility (n = 12k, that is, n is a multiple of 12), n^2+n = (12k)^2+(12k) = 144k^2+12k = 12(12k^2+k) But, n^2+n = 12g, so this case gives: 12(k^2+k) = 12g, that is, g = 12k^2+k The second possibility (n = 12k+3, that is, n is 3 more than a multiple of 12) similarly gives n^2+n = (12k+3)^2+(12k+3) = 144k^2+84k+12 = 12(12k^2+7k+1) 12(12k^2+7k+1) = 12g, so, g = 12k^2+7k+1 The other possibilities (n = 12k+8, n=12k+11) give after similar calculations, g = 12k^2+17k+6, and g = 12k^2+23k+11 If k > 0, then all four possibilities clearly give nonnegative values for g. (For k = 0, we generate the first four values of your sequence: 0, 1, 6, 11.) Suppose we want to generate a list of such values: We can exhaust all the possibilities for g by evaluating the above four formulas for g for each k = 0,1,2,3,... (When writing your list, because you need positive integer values for g, you'll need to throw away the first element of your list, 12(0)^2+(0)=0.) Considering the values of g generated by a particular k, we have (clearly): 12k^2+k < 12k^2+7k+1 < 12k^2+17k+6 < 12k^2+23k+11. So, the smallest value of g generated by k+1 is: 12(k+1)^2+(k+1) = 12k^2+25k+13 > 12k^2+23k+11 That is, the smallest value for g generated by k+1 is larger than the largest value generated by k. So, we can write down a complete ordered list (a sequence that gives the possible solutions for g in increasing order) by writing down the four possibilities for g generated by 0 (in increasing order), then the four generated by 1, then those for 2, and so on. I don't know how to write the sequence more compactly than by writing down the above rule. Travis > Hi all, > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? > cdj === Subject: Integral Hi. What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral >Hi. >What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? http://integrals.wolfram.com/ says: RootSum[-1+#1+#1^2+#1^5 &, (Log[x-#1] / (1+2#1+5#1^4)) &] Thomas === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? As soon as you come up with a closed form of factorization of x^5 + x^2 + x - 1 (one linear factor and two quadratic factors), I will be able to tell you more. === Subject: Re: Integral Mike4ty, Ugly, I think. Factor the polynomial (warning, it's irreducible over the integers), then apply the method of partial fractions. Travis > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? The hard part is solving the quintic. There is a real root near 0.568544, a complex conjugate pair near 0.91612 +/- 0.57771 i, and another pair near 0.622848 +/- 1.03222 i. Once you have the factors, it's an easy partial fractions decomposition, provided you don't mind approximate answers. Mathematica 5.0 for Mac OS X -- Terminal graphics initialized -- In[1]:= Integrate[1./(x^5+x^2+x-1),x] Out[1]= 1. (-0.22894 ArcTan[0.484391 (-1.2457 + 2. x)] - > 0.189874 ArcTan[0.865487 (1.83224 + 2. x)] + > 0.361679 Log[0.586544 - 1. x] - 2 > 0.0641575 Log[1.45342 - 1.2457 x + x ] - 2 > 0.116682 Log[1.17302 + 1.83224 x + x ]) -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Galois group = A_4 The following question has been bugging me : I am trying to come up with a quartic polynomial whose Galois group is A_4. I know I can go about the business of finding a discriminant that is the square of a rational number, and the resolvent cubic is irreducible. But I was wondering if there was a more illuminating way to geometrically come up with a quartic polynomial. In particular, we know A_4 has no transpositions and no 4-cycles. So, what can I say about the quartic? Wouldn't it be true that since there are no transpositions, the quartic cannot have exactly 2 real roots? What can I deduce about the fact that the Galois group has no 4-cycles? Tony === Subject: Re: How to find the extremum of the Absolute value of a function=?big5?Q?=EF=BC=9F?= My method is to differentiate |Z| w.r.t y, and then subtitute y1 and y2 into the above equation. Finally I got a simultaneous equations and solving for x1 and x2. But where confused me is that the derivative of an absolute value seems not exist, so my known method didn't work due to the absolute value. is there any other method to solve a extremum problem of an obsolute value? Á¡ (Randy Poe)ÁnÛ¤Èæ2 50´ÁG > Suppose I have a complex-valued function Z, and Z=3DZ(x1,x2,y) > where x1,x2 and y are three real variables. I wanna |Z| has local > minimums at two given points y=3Dy1 and y=3Dy2, where x1 and x2 should be > adjusted to met this demand. > i.e. > Q=EF=BC=9AHow to find x1 and x2 such that |Z| has local minimum at two > given points > y1 and y2 ? > To find the minimimum for y=3Dy1, define a new function: > W1(x1,x2) =3D Z(x1, x2, y1) > and use your favorite minimization technique. Similarly for y2. > - Randy -- Á¡ Origin: ´.beÛjË.b3.bc®[UDo ubleDot].90[Degre e]Tøü  === Subject: swjpam fyi, The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer exists due to budgetary contraints. === Subject: Re: swjpam Discussion, linux) > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. I will never doubt the hammer again. Golly. -- Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others. -- Ambrose Bierce === Subject: Re: swjpam > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. Is that just the excuse? Is the real reason editorial incompetence? Is this the first blow of The Hammer? === Subject: do you have any smart way of finding which number is bigger ? Using the fatest way: compare: 0.9^10 vs. 2*(0.9^19)+0.9^20 how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? lucy escribi.97: > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? 2*(0.9^19)+0.9^20 = 0.9^10(2*0.9^9 + 0.9^10) = = 0.9^10(20*9^9 + 9^10)/10^10 = 0.9^10* 9^9*29/10^10 But 9^2 = 81 > 80, then 9^8 > 8^4*10^4 = 4096*10^4 ==> 9^9 > 36*10^7 ==> 29*9^9 > 36*29*10^7 = 1044*10^7 > 10^10 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to over 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? A: 0.9^10 B: 2*(0.9^19)+0.9^20 Notice both pieces of B are positive. b1 > 0 b2>0 So if I can see one side of + larger than A then B > A 0.9^20 > 0.9^10 20 > 10 ergo B > A 5 seconds === Subject: Re: do you have any smart way of finding which number is bigger ? lucy, Here's a way that doesn't use much explicit calculation. (It admittedly uses the fact that e < 2.9, which can easily be derived analytically, anyway). The Taylor series for log about 1 is: log(1+x) = x - (1/2)x^2 + (1/3)x^3 - ... Setting x = 1/9, we have: log(10/9) = log(1 + 1/9) = (1/9) - (1/2)(1/9)^2 + (1/3)(1/9)^3 - ... < 1/9 9 * log (10/9) < 1 9 log 10 - 9 log 9 < 1 Adding log 10, 10 log 10 - 9 log 9 < 1 + log 10 = log (e * 10) < log (29) [Here is where I invoke e < 2.9.] Subtracting log 9, 10 log 10 - 10 log 9 < log 29 - log 9 10 log (10/9) < log (29/9) = log (2 * (10/9) + 1) Exponentiating, (10/9)^10 < 2*(10/9) + 1 Multiplying both sides by (9/10)^20 (9/10)^10 < 2*(9/10)^19 + (9/10)^20 I suspect there is a much more elegant way. Maybe something with the expression 10 log 10 - 9 log 9 < log (29)? Travis > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? I don't know if it's smart or fastest, but you can factor. 0.9^10 >?< 2 * (0.9^19) + 0.9^20 0.9^10 >?< 2 * (0.9^20)/0.9 + 0.9^20 0.9^10 >?< 0.9^20 * (2/0.9 + 1) 0.9^10 / 0.9^20 >?< 2/0.9 + 1 0.9^-10 >?< 2.2222... + 1 2.8679... < 3.2222... -- john === Subject: Re: do you have any smart way of finding which number is bigger ? ETAtAhUAvgUxhHwS5c+oabk25UmVuHI06JUCFAza+gLCc85dGRk3KQeD5aQy6jyx 0.9^10 ? 2*(0.9^19)+0.9^20 1 ? 2^(0.9^9) + 0.9^10 10^10 ? 20*9^9 + 9^10 = 29*9^9 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. Therefore ? is <. --OL === Subject: Re: do you have any smart way of finding which number is bigger ? > 0.9^10 ? 2*(0.9^19)+0.9^20 > 1 ? 2^(0.9^9) + 0.9^10 > 10^10 ? 20*9^9 + 9^10 = 29*9^9 I followed you up to this point; you are now comparing 10^10 with 29 * 9^9 I don't understand the next line. > 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. > Therefore ? is <. -- john === Subject: Re: do you have any smart way of finding which number is bigger ? .9^10 = 0.3486784401 2*(0.9^19) + 0.9^20 = 0.39174699812516770581 The second one is larger time to do the problem -- (2 seconds for cut and paste maybe?) === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I don't see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you don't believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you don't believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). --Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! you're not your Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didn't know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didn't know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Here's a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is r's n'th digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but that's not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker What's a hyper-real number? Do you even know anything about math? Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> What's a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least it's a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that I'm all that neat, but hopefully my notation's clear at least.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You don't even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You don't even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, don't look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. I'll define it with an example. Suppose you have alot of people here making noise here on this NG and they don't know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You don't even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, don't look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > I'll define it with an example. Suppose you have alot of people here >making >noise here on this NG and they don't know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry don't look... Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still don't know what a hyper-real number is. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: terms in Math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmj09356; >Can anyone tell me the URL with pages devoted to math term's. One such is http://mathworld.wolfram.com/ === Subject: Re: coverings of a Mobius band by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmT09372; >Are the even degree covers of a Mobius band annulus(es)? Are the odd degree covers of a Mobius band, Mobius bands again? What is the universal cover of a Mobius band? Yes, yes, and R x [-1, 1]. One can imagine a strip of paper with n half-twists before gluing the ends as the total space of the degree n cover of the Mobius band. The strip of paper is I x [-1, 1], and the gluing map for taping the ends is the homeomorphism (-1)^n * - : [-1, 1] --> [-1, 1], so we get an annulus for n even and a Mobius strip for n odd. For the universal cover, imagine an infinite strip with infinitely many half-twists. Todd Trimble === Subject: Re: The Possible Cure For AIDS in part: > There are no A's, G's, 4's, O's, >>The G is gimmel the 3 rd letter of the Hebrew alphabet. >>The vowels (mplicit in old Hebrew but made explicit by the Masorites) >>include an ah sound and a oh sound. >AFAIK the Torah codes are never done with the vowels. That's quite true. Including vowels makes it harder to find matches. But if someone wanted to find English words in the Tanakh, the standard convention is, of course, Aleph for A and Ayin for O. As for the number 404, two letters would represent that in standard Hebrew numeric notation: Tau and Daleth. But Hebrew has a word for silver, although *ancient* Hebrew hardly had a word for oxygen, even if it certainly has one today. Actually, we should be asking what letters of the Hebrew alphabet are supposed to stand for and ! John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Possible Cure For AIDS > But Hebrew has a word for silver, although *ancient* Hebrew hardly had a Kessef. > word for oxygen, even if it certainly has one today. Khamtzan which means sour-er. This corresponds to the German Saurstuff. > Actually, we should be asking what letters of the Hebrew alphabet are > supposed to stand for and ! There is a typefont used to crossreference passages in the Talmud to the Shulkhan Arooch. It looks like superscripts because of the way it is printed. This was originated about 1500 of the common era. Bob Kolker === Subject: Re: The Possible Cure For AIDS in part: >There are no A's, G's, 4's, O's, etc. in Hebrew. Did you transliterate >these according to some arbitrary scheme, or did you search for Hebrew >letter strings? Not that it matters, of course. But while we're talking about a cure for AIDS... Some years back, there were news stories about how, for AIDS research, modified mice were created with human immune systems. This was done by taking cells from human fetuses. Naturally, this was controversial because of the abortion issue. I noted that nobody cares about mouse fetuses. If you can fix a mouse so that it _can_ get AIDS by giving it a human being's immune system, it would seem you could fix a human so that he can't get AIDS by giving him a mouse's immune system. There were probably very good reasons why this couldn't _really_ be done, but I had thought it worth mentioning. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Beal's conjecture > => A large prize is offered by banker Beal for a solution to > => the Beal Conjecture: the equation x^p + y^q = z^r has no solutions > => for p, q, r > 2 and coprime integers x, y, z. > = => Sorry if this has been discussed here - my only justification > => is that I just recently discovered this group but I am curious to > => know if Wile's proof of FLT also covers the Beal Conjecture > => Or have any counter-examples been found? Wile's proof covered the case p=q=r and relaxed the coprime requirement for x, y, & z. IE: There is no integer solution when p=q=r whatever the coprimality of x, y, & z. tom -- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an hour. === Subject: Mathforge.net :: Near-numbers, autonomic computing, John Derbyshire, and Church-Turing The Latest Math News from Mathforge.net http://mathforge.net ****An invitation to additive prime number theory**** A.V. Kumchev and D.I. Tolev have compiled a short document entitled An Invitation to Additive Prime Number Theory[~60pp, pdf]. The document serves as an introductory guide to graduate-level students on... http://mathforge.net/index.jsp?page=seeReplies&messageNum=981 ****Are there encoded messages in the Bible?**** researchers both supporting and denying the statistical evidence for 'hidden messages' found in the Old Testament tell their tales. Some... http://mathforge.net/index.jsp?page=seeReplies&messageNum=974 ****Near-numbers: the new 'limit'**** There's a very interesting paper by Frank J Swenton of Middlebury College called Limits and the System of Near-Numbers[19pp, pdf]. What seemed at first glance (at the title and abstract) like an uninspired... http://mathforge.net/index.jsp?page=seeReplies&messageNum=970 ****Introduction to autonomic computing**** While not itself mathematical in nature, the concept is built around ideas garnered from areas of artificial intelligence research and it is easy to see that autonomic computing could have applications... http://mathforge.net/index.jsp?page=seeReplies&messageNum=969 ****Derbyshire's Diary**** John Derbyshire, author of Prime Obsession, has a mathematical problem accompanying each of his Diary entries located in his Web Journalism folder. If you sift through enough of the partisan propaganda... http://mathforge.net/index.jsp?page=seeReplies&messageNum=968 ****Maple 9.5 Released**** MapleSoft has released version 9.5 of their popular symbolic and numeric computational software suite Maple. New Features include added packages (optimization, logic, and root finding), OpenMaple access... http://mathforge.net/index.jsp?page=seeReplies&messageNum=967 ****Mathematica 5.1 released**** The Mathematica version has jumped a tenth of a point, and Wolfram has added scores of new features to the new release, including Web Services support, a benchmarking package, string manipulation functions,... http://mathforge.net/index.jsp?page=seeReplies&messageNum=966 ****Quantum computers and the Church-Turing Thesis**** The original Church-Turing Thesis states that every function which would naturally be regarded as computable can be computed by a Turing Machine and Petrus H. Potgeiter mentions in his paper Zeno Machines... http://mathforge.net/index.jsp?page=seeReplies&messageNum=963 ****Sobering U.S. Student Math Scores**** In a disheartening follow-up to the Putnam story below, news (Seattle Times) outlets (New York Times)everywhere are reporting the horrid state of U.S. student math skills. The Organisation for Economic... http://mathforge.net/index.jsp?page=seeReplies&messageNum=959 ''Mathforge ran a story about the Putnam Competition... http://mathforge.net/index.jsp?page=seeReplies&messageNum=958 === Subject: What kind of matrix can map positive element vectors to positive element vectors? Hi all, Suppose I have a matrix A, and a positive element vector x, then y=A*x, I want y to be also positive element vector... What can I say about A? I want all such kinds of A? What kinds of A can let me have both directions: x positive elements <=> y=A*x positive elements? === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > Suppose I have a matrix A, and a positive element vector x, > then y=A*x, > I want y to be also positive element vector... > What can I say about A? I want all such kinds of A? > What kinds of A can let me have both directions: > x positive elements <=> y=A*x positive elements? Take a simple case and try to learn from it. The unit vectors for x will pull out the columns of A. Sure seems to say that A must at least have all of its elements positive. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > > Suppose I have a matrix A, and a positive element vector x, > > then y=A*x, > > I want y to be also positive element vector... > > What can I say about A? I want all such kinds of A? > > What kinds of A can let me have both directions: > > x positive elements <=> y=A*x positive elements? > > > > Take a simple case and try to learn from it. > The unit vectors for x will pull out the columns of A. > Sure seems to say that A must at least have all of its > elements positive. Necessary and sufficient. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? >> Hi all, >> Suppose I have a matrix A, and a positive element vector x, >> then y=A*x, >> I want y to be also positive element vector... >> What can I say about A? I want all such kinds of A? >> What kinds of A can let me have both directions: >> x positive elements <=> y=A*x positive elements? >> Take a simple case and try to learn from it. >> The unit vectors for x will pull out the columns of A. >> Sure seems to say that A must at least have all of its >> elements positive. >Necessary and sufficient. Neither necessary (unless your positive means >= 0) nor sufficient. Note that lucy wanted <=>. I'll assume A is a square matrix. It's easy to see A must be nonsingular, else given vector x > 0 (i.e. all x_i > 0) with Ax > 0, you could add to x a suitable multiple of a vector w with Aw = 0 to get a vector v not > 0 with Av = Ax > 0. A must map the nonnegative cone C = {x in R^n: all x_i >= 0} onto itself. Note that the extreme rays of C pass through the standard unit vectors e(j) (with e(j)_i = 1 if i=j, 0 otherwise). That is, these are the only members w of C such that if w = t x + (1-t) y with x,y in C and 0 < t < 1, then x and y are scalar multiples of w. Now A must map extreme rays to extreme rays, and from this it's easy to see that A must be of the form A = D P where P is a permutation matrix and D a diagonal matrix with positive elements on the diagonal. Conversely, everything of this form has the desired property. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: How to determine parameter integer values such that quadratic has integer solutions? Hi all, In particular, I'm interested in finding the positive integer values of g such that the following quadratic in v has integer roots: v^2 - 7v + (12 - 12g) Evidently this requires that 1 + 48g be a perfect square. I can have Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, 20, 35, ... This sequence is more conveniently summarized as something-mod-something-else, but I forget exactly what the two somethings are. How to figure this out? cdj === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) v = (7 +- sqr(49 - 48 + 48g)/2 = (7 +- sqr(1 + 48g)/2 sqr(1 + 48g) must be odd number, 2n + 1 1 + 48g = 4n^2 + 4n + 1 12g = n^2 + n = n(n + 1) Case 12 | n. Done Case 4 | n, not 3 | n. 3 | n+1; n = 3k - 1 Case 3 | n, not 2 | n. 4 | n+1; n = 4k - 1 Case 2 | n, not 4 | n. Not possible Three family parametrization of solutions. n = 12k n = 3k - 1 provided 4 | n n = 4k - 1 provided 3 | n Expect some duplicates. > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? The square root of 1+48g is odd... for fuck's sake I'm blind.... === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? cdj: Here's an outline: We know that 1+48g is a perfect square. Since g is an integer, 48g is even, and 1+48g is odd. So, the square root of 1+48g is odd, and we can write that square root as 2n+1 for some integer n. Thus, (2n+1)^2 = 1+48g Expanding and simplifying, 4n^2+4n+1 = 1+48g 4n^2+4n = 48g n^2+n = 12g We need a little number theory here: Consider the above equation (mod 12): (n^2+n) (mod 12) = 0 Trying the 11 possibilities, we find that n (mod 12) must be 0, 3, 8, or 11, that is, we must be able to write n as 12k or 12k+3 or 12k+8 or 12k+11 for some integer k. (The above condition is equivalent to (n^2+n) = 0 (mod 3) AND (n^2+n) = 0 (mod 4), so with some work, we really need only to consider 3+4 = 7 cases.) Substituting for the first possibility (n = 12k, that is, n is a multiple of 12), n^2+n = (12k)^2+(12k) = 144k^2+12k = 12(12k^2+k) But, n^2+n = 12g, so this case gives: 12(k^2+k) = 12g, that is, g = 12k^2+k The second possibility (n = 12k+3, that is, n is 3 more than a multiple of 12) similarly gives n^2+n = (12k+3)^2+(12k+3) = 144k^2+84k+12 = 12(12k^2+7k+1) 12(12k^2+7k+1) = 12g, so, g = 12k^2+7k+1 The other possibilities (n = 12k+8, n=12k+11) give after similar calculations, g = 12k^2+17k+6, and g = 12k^2+23k+11 If k > 0, then all four possibilities clearly give nonnegative values for g. (For k = 0, we generate the first four values of your sequence: 0, 1, 6, 11.) Suppose we want to generate a list of such values: We can exhaust all the possibilities for g by evaluating the above four formulas for g for each k = 0,1,2,3,... (When writing your list, because you need positive integer values for g, you'll need to throw away the first element of your list, 12(0)^2+(0)=0.) Considering the values of g generated by a particular k, we have (clearly): 12k^2+k < 12k^2+7k+1 < 12k^2+17k+6 < 12k^2+23k+11. So, the smallest value of g generated by k+1 is: 12(k+1)^2+(k+1) = 12k^2+25k+13 > 12k^2+23k+11 That is, the smallest value for g generated by k+1 is larger than the largest value generated by k. So, we can write down a complete ordered list (a sequence that gives the possible solutions for g in increasing order) by writing down the four possibilities for g generated by 0 (in increasing order), then the four generated by 1, then those for 2, and so on. I don't know how to write the sequence more compactly than by writing down the above rule. Travis > Hi all, > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? > cdj === Subject: Integral Hi. What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral >Hi. >What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? http://integrals.wolfram.com/ says: RootSum[-1+#1+#1^2+#1^5 &, (Log[x-#1] / (1+2#1+5#1^4)) &] Thomas === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? As soon as you come up with a closed form of factorization of x^5 + x^2 + x - 1 (one linear factor and two quadratic factors), I will be able to tell you more. === Subject: Re: Integral Mike4ty, Ugly, I think. Factor the polynomial (warning, it's irreducible over the integers), then apply the method of partial fractions. Travis > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? The hard part is solving the quintic. There is a real root near 0.568544, a complex conjugate pair near 0.91612 +/- 0.57771 i, and another pair near 0.622848 +/- 1.03222 i. Once you have the factors, it's an easy partial fractions decomposition, provided you don't mind approximate answers. Mathematica 5.0 for Mac OS X -- Terminal graphics initialized -- In[1]:= Integrate[1./(x^5+x^2+x-1),x] Out[1]= 1. (-0.22894 ArcTan[0.484391 (-1.2457 + 2. x)] - > 0.189874 ArcTan[0.865487 (1.83224 + 2. x)] + > 0.361679 Log[0.586544 - 1. x] - 2 > 0.0641575 Log[1.45342 - 1.2457 x + x ] - 2 > 0.116682 Log[1.17302 + 1.83224 x + x ]) -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Galois group = A_4 The following question has been bugging me : I am trying to come up with a quartic polynomial whose Galois group is A_4. I know I can go about the business of finding a discriminant that is the square of a rational number, and the resolvent cubic is irreducible. But I was wondering if there was a more illuminating way to geometrically come up with a quartic polynomial. In particular, we know A_4 has no transpositions and no 4-cycles. So, what can I say about the quartic? Wouldn't it be true that since there are no transpositions, the quartic cannot have exactly 2 real roots? What can I deduce about the fact that the Galois group has no 4-cycles? Tony === Subject: Re: How to find the extremum of the Absolute value of a function=?big5?Q?=EF=BC=9F?= My method is to differentiate |Z| w.r.t y, and then subtitute y1 and y2 into the above equation. Finally I got a simultaneous equations and solving for x1 and x2. But where confused me is that the derivative of an absolute value seems not exist, so my known method didn't work due to the absolute value. is there any other method to solve a extremum problem of an obsolute value? Á¡ (Randy Poe)ÁnÛ¤Èæ2 50´ÁG > Suppose I have a complex-valued function Z, and Z=3DZ(x1,x2,y) > where x1,x2 and y are three real variables. I wanna |Z| has local > minimums at two given points y=3Dy1 and y=3Dy2, where x1 and x2 should be > adjusted to met this demand. > i.e. > Q=EF=BC=9AHow to find x1 and x2 such that |Z| has local minimum at two > given points > y1 and y2 ? > To find the minimimum for y=3Dy1, define a new function: > W1(x1,x2) =3D Z(x1, x2, y1) > and use your favorite minimization technique. Similarly for y2. > - Randy -- Á¡ Origin: ´.beÛjË.b3.bc®[UDo ubleDot].90[Degre e]Tøü  === Subject: swjpam fyi, The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer exists due to budgetary contraints. === Subject: Re: swjpam Discussion, linux) > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. I will never doubt the hammer again. Golly. -- Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others. -- Ambrose Bierce === Subject: Re: swjpam > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. Is that just the excuse? Is the real reason editorial incompetence? Is this the first blow of The Hammer? === Subject: do you have any smart way of finding which number is bigger ? Using the fatest way: compare: 0.9^10 vs. 2*(0.9^19)+0.9^20 how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? lucy escribi.97: > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? 2*(0.9^19)+0.9^20 = 0.9^10(2*0.9^9 + 0.9^10) = = 0.9^10(20*9^9 + 9^10)/10^10 = 0.9^10* 9^9*29/10^10 But 9^2 = 81 > 80, then 9^8 > 8^4*10^4 = 4096*10^4 ==> 9^9 > 36*10^7 ==> 29*9^9 > 36*29*10^7 = 1044*10^7 > 10^10 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to over 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? A: 0.9^10 B: 2*(0.9^19)+0.9^20 Notice both pieces of B are positive. b1 > 0 b2>0 So if I can see one side of + larger than A then B > A 0.9^20 > 0.9^10 20 > 10 ergo B > A 5 seconds === Subject: Re: do you have any smart way of finding which number is bigger ? lucy, Here's a way that doesn't use much explicit calculation. (It admittedly uses the fact that e < 2.9, which can easily be derived analytically, anyway). The Taylor series for log about 1 is: log(1+x) = x - (1/2)x^2 + (1/3)x^3 - ... Setting x = 1/9, we have: log(10/9) = log(1 + 1/9) = (1/9) - (1/2)(1/9)^2 + (1/3)(1/9)^3 - ... < 1/9 9 * log (10/9) < 1 9 log 10 - 9 log 9 < 1 Adding log 10, 10 log 10 - 9 log 9 < 1 + log 10 = log (e * 10) < log (29) [Here is where I invoke e < 2.9.] Subtracting log 9, 10 log 10 - 10 log 9 < log 29 - log 9 10 log (10/9) < log (29/9) = log (2 * (10/9) + 1) Exponentiating, (10/9)^10 < 2*(10/9) + 1 Multiplying both sides by (9/10)^20 (9/10)^10 < 2*(9/10)^19 + (9/10)^20 I suspect there is a much more elegant way. Maybe something with the expression 10 log 10 - 9 log 9 < log (29)? Travis > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? I don't know if it's smart or fastest, but you can factor. 0.9^10 >?< 2 * (0.9^19) + 0.9^20 0.9^10 >?< 2 * (0.9^20)/0.9 + 0.9^20 0.9^10 >?< 0.9^20 * (2/0.9 + 1) 0.9^10 / 0.9^20 >?< 2/0.9 + 1 0.9^-10 >?< 2.2222... + 1 2.8679... < 3.2222... -- john === Subject: Re: do you have any smart way of finding which number is bigger ? ETAtAhUAvgUxhHwS5c+oabk25UmVuHI06JUCFAza+gLCc85dGRk3KQeD5aQy6jyx 0.9^10 ? 2*(0.9^19)+0.9^20 1 ? 2^(0.9^9) + 0.9^10 10^10 ? 20*9^9 + 9^10 = 29*9^9 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. Therefore ? is <. --OL === Subject: Re: do you have any smart way of finding which number is bigger ? > 0.9^10 ? 2*(0.9^19)+0.9^20 > 1 ? 2^(0.9^9) + 0.9^10 > 10^10 ? 20*9^9 + 9^10 = 29*9^9 I followed you up to this point; you are now comparing 10^10 with 29 * 9^9 I don't understand the next line. > 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. > Therefore ? is <. -- john === Subject: Re: do you have any smart way of finding which number is bigger ? .9^10 = 0.3486784401 2*(0.9^19) + 0.9^20 = 0.39174699812516770581 The second one is larger time to do the problem -- (2 seconds for cut and paste maybe?) === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I don't see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you don't believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you don't believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). --Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! you're not your Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didn't know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didn't know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Here's a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is r's n'th digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but that's not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker What's a hyper-real number? Do you even know anything about math? Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> What's a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least it's a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that I'm all that neat, but hopefully my notation's clear at least.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You don't even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You don't even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, don't look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. I'll define it with an example. Suppose you have alot of people here making noise here on this NG and they don't know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You don't even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, don't look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > I'll define it with an example. Suppose you have alot of people here >making >noise here on this NG and they don't know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry don't look... Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still don't know what a hyper-real number is. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: hyper-ellipse Epigone-thread: gleezhimzhal Is there a proper equation for confocal n-dimensional ellipses and m-dimensional hyperbolas in n+m dimensions, where n=m? Since the elliptic surfaces are normal to the hyperbolic surfaces, is it possible to represent them as surfaces in different subspaces of n+m dimensions? === Subject: Re: Banach space Of Analytic Functions >>Of Course I Search for a NON STANDARD topology for some Functional >>Space(For Example schwars maps) which is invariant under the action >>a polynomial vector field X (As a Bounded operator ) >In a certain sense, such spaces have *only one* natural topology: >complete, separable, metrizable, and described without resort to the >Axiom of Choice. But I guess it takes a logician to specify what that >sense is. Someone might point out that the standard topology on the Schwarz space _is_ invariant under the action of a polynomial vector field (as a bounded operator) - it's just not given by a norm. ************************ David C. Ullrich === Subject: A Question on Control theory Epigone-thread: himpspaybang Is Commutativity of Lie algebra A below the weakest condition for wich the following statment is valid: Let A be a commutative lie Algebra of Vector fields on a manifold M, assume an element of A , say X, has a periodic attractor. Then the System A is not Controlable *for Definition Of Periodic Attractor see Hirsch Smale Linear Algebra Diff Eq. Dynamical System Can One replace Commutativity of A with another assumption? === Subject: Re: Why exp(-st) in the Laplace Transform? <7c3wd.1100621$Gx4.435402@bgtnsc04-news.ops.worldnet.att.net Paul, > I am not sure that I comprehend all of your commentary, however I do > note one of your statements > If you take the Laplace transform, ... > Shifting an exponential (exp(-st)) by a constant (t -> t + a) is > equivalent to multiplying it by a real number (exp(-sa)). > Your statement is not quite valid, in that, for the case of a Laplace > transform, the transform variable s has both real and imaginary > components, thus exp(-sa) has both real and imaginary components as > well. Terminology issue. In many fields a Laplace transform means exclusively real-valued exponent. If it's complex than it's a complex Laplace transform or even generalized Laplace transform, though that's also used for other extensions, such as adding discrete portions. > As in the Fourier transform, you would integrate over the > imaginary component. It is the real component, however, which > distinguishes the Laplace Transform from the Fourier Transform. That > is, the real component of our multiplier will tend to attenuate the > function. An example of the use of a Laplace transform would be to > transform the unit step function U(t). Note that it's Laplace Transform > is determined to be 1/s, under the restriction re(s) > 0. The Fourier > Transform of U(t) does not exist. Well, that depends on what you mean by exist. It sure does, if you allow distributions, rather than just functions as Fourier transforms. Unsurprisingly, the non-singular portion is -i/k, or what you get for 1/s, when you replace s with (ik). The singular portion is just a delta distribution, caused by the (1/2) offset that is the even portion of the unit step. One way to do this is to take a series of function that converge to the unit step but whose Fourier transforms are functions -- e.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving the delta distribution noticed above. Write signum(x) as the limit a -> 0 of f_a(t) = e^{-at} for t > 0, -e^{at} for t < 0. The fourier transform of these will be -1/(a - ik) + 1/(a + ik) = (-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a -> 0 = 2/ki, but we need to divide by 2 to get the step function. Physicists often just use it via formal manipulation in integrals, which while not normally made rigorous, does give workable results, along the lines of: The unit step is just the integral of the delta function. Integrating a function in the real domain can be done by dividing the function in the Fourier domain by ik. Again this gives the non-singular portion as 1/ik. I have left out constant multiplicative factors, as they'll depend on one's convention. -- Aaron Denney -><- === Subject: Re: Why exp(-st) in the Laplace Transform? <7c3wd.1100621$Gx4.435402@bgtnsc04-news.ops.worldnet.att.net> Aaron, Your point is well made. Indeed, the Laplace transform of U(t) is determined by the real part of s, the transform variable corresponding to t. So, we would have Laplace transform of U(t) is 1/s for re(s) > 0 Fourier transform of U(t) is pi*delta(k) + 1/ik for the case re(s) = 0. s, in this case, being equal to ik. As you point out, each piece of the signum function which you define is Fourier transformable. Note that your selection of exp(-at) for the U(t) portion is, in essence, adding the positive real component a to the Fourier transform variable ik. That is, lim a->0 of Fourier transform of exp(-at)U(t) is identical to the limit re(s)->0 of Laplace transform of U(t), having re(s) > 0. A similar argument is made for the U(-t) portion. One could therefore argue that the Fourier transform of U(t) exists, since its Laplace transform exists in the limiting sense as re(s)->0. Ed -- Edward Hyman EdwardH@email.uophx.edu Other EMail: e.hyman@worldnet.att.net > Paul, > I am not sure that I comprehend all of your commentary, however I do > note one of your statements > If you take the Laplace transform, ... > Shifting an exponential (exp(-st)) by a constant (t -> t + a) is > equivalent to multiplying it by a real number (exp(-sa)). > Your statement is not quite valid, in that, for the case of a Laplace > transform, the transform variable s has both real and imaginary > components, thus exp(-sa) has both real and imaginary components as > well. Terminology issue. In many fields a Laplace transform means exclusively real-valued exponent. If it's complex than it's a complex Laplace transform or even generalized Laplace transform, though that's also used for other extensions, such as adding discrete portions. > As in the Fourier transform, you would integrate over the > imaginary component. It is the real component, however, which > distinguishes the Laplace Transform from the Fourier Transform. That > is, the real component of our multiplier will tend to attenuate the > function. An example of the use of a Laplace transform would be to > transform the unit step function U(t). Note that it's Laplace Transform > is determined to be 1/s, under the restriction re(s) > 0. The Fourier > Transform of U(t) does not exist. Well, that depends on what you mean by exist. It sure does, if you allow distributions, rather than just functions as Fourier transforms. Unsurprisingly, the non-singular portion is -i/k, or what you get for 1/s, when you replace s with (ik). The singular portion is just a delta distribution, caused by the (1/2) offset that is the even portion of the unit step. One way to do this is to take a series of function that converge to the unit step but whose Fourier transforms are functions -- e.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving the delta distribution noticed above. Write signum(x) as the limit a -> 0 of f_a(t) = e^{-at} for t > 0, -e^{at} for t < 0. The fourier transform of these will be -1/(a - ik) + 1/(a + ik) (-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a -> 0 = 2/ki, but we need to divide by 2 to get the step function. Physicists often just use it via formal manipulation in integrals, which while not normally made rigorous, does give workable results, along the lines of: The unit step is just the integral of the delta function. Integrating a function in the real domain can be done by dividing the function in the Fourier domain by ik. Again this gives the non-singular portion as 1/ik. I have left out constant multiplicative factors, as they'll depend on one's convention. -- Aaron Denney -><- === Subject: Sign of the remainder of Taylor expansion Epigone-thread: dwaxthayquyr Let f(x) be a real valued function, with Taylor expansion (around zero) : f(x) = sum_{k=0}^{infty} (1/k!) * (df^k(0) / dx^k) * x^k If we take only M+1 coefficients, we get : f(x) = sum_{k=0}^{M} (1/k!) * (df^k(0) / dx^k) * x^k + R_M(x) Where R_M is the Lagrange remainder. So, if we're inside the radius of convergence, we get that R_M(x) --> 0 when M --> infty. Are there any known condition for this convergance to be always from 'above' ? that is, that R_M(x) < 0 for all M ? (even for a specific x) Note that this is weaker then demanding monotone convergance. (My example for f is some complicated rational function, for which I'm trying to prove this, and I don't have a general formula for the k'th derivative) Thanx, Dan === Subject: Galois group of a given quartic equation, part II Epigone-thread: twomprendlex Another wellknown ladder problem, the so-called CLP (crossed ladders problem) is also connected to a quartic equation Q(x) = 0. For a, b and c integers and K = a^2 - b^2, what can be said about the Galois group G of the quartic equation Q(x) = x^4 - 2 c x^3 - K x^2 + 2 c K x - K c^2 = 0? For the sake of determining G we can assume that a>b>0,c>0 and GCD[a,b,c]=1. Let m be the number of integral roots of the equation Q(x) = 0 (counting roots with multiplicity). It can be shown that the discriminant of the quartic equation is always negative. The discriminant is therefore never a square. The equation always has 2 real and 2 complex (conjugate) roots. The Galois group is therefore nonabelian in the irreducible case, see http://mathforum.org/epigone/sci.math.research/yecholloi). The following is then easily proved: I. In the case Q(x) is irreducible G is either S4 or D4. II. In the case m = 1, G is S3. III. In the case m = 2, G is Z2. IV. In the case Q(x) is reducible and m = 0, G is V = Z2 x Z2. (Note that we cannot have m = 4 since then the discriminant must be a square, so the trivial Galois group cannot occur.) It is easy to see that if Q(x) is irreducible and K = c^2 (i.e. a^2 = b^2 + c^2, (a,b,c) a Pythagorean triple) then G is D4. A. It is unknown (at least to me) if there are examples of the cases III. and IV. B. It is also unknown (at least to me) if there are cases with G = D4 for K /= c^2. Can any of the readers help with A. and B.? Mathematical Gazette.) Kent Holing === Subject: CFP - Benelux Information Theory 26th Symposium on INFORMATION THEORY in the BENELUX May 19th-May 20th, 2005, Brussels, Belgium Universitò Libre de Bruxelles http://www.ulb.ac.be/di/benelux05 organized under the auspices of Werkgemeenschap voor Informatie- en Communicatietheorie (www.w-i-c.org) === Subject: Re: Uniqueness of implicit functions Epigone-thread: zhoabrarshel I want to bring into 3 ideas: -In constant functions g(x1,x2,;.,xi,..xn)=c g:R^n->{c} g may be broken many a way : 1)separating variables: g() -> h1(x1)=h2(x2,x3,..xn) ;h(xn)=h2(x1,x2,..xn-1) l(x1,x3,x4)=m(x2,x5,x6...xn) Example : x1^2+x2^2-x3^2=0 ->x3^2-x1^2=x2^2 2)repeating variables: g() -> p(x1,x2,x3)=q(x1,x2,x3), sphere x1^2+x2^2+x3^2= R^2 , x1^2+x2^2-2x2+x3= R^2-2x2-x^3+x3 ... -Implicit functions (c=0) may be built from a only implicit function: f1(x1,x2,;.,xi,..xn)=0 let us choose pi invertible,defined at 0 , f2() =p1(f1( ))-p1(0), f3() =p2(f2( ))-p2(0), or f4( , )= p3(f1,f2)-p3(0,0) pj:R^2->R , f5( , , ) = p4(f1,f2,f3)-p4(0,0,0) p4:R^3->R . -Iterated forms for constant functions if g(x,y) = f^[m(x,y)] n(x,y)= c constant, n(x,y) continuous number of iteration of map f then: n(x,y)=f^[-m(x,y)] c , Alain. === Subject: This week in the mathematics arXiv (13 Dec - 17 Dec) Here are this week's titles in the mathematics arXiv, available at: http://front.math.ucdavis.edu/ http://front.math.ucdavis.edu/submissions This week in the mathematics arXiv may be freely redistributed with attribution and without modification. Titles in the mathematics arXiv (13 Dec - 17 Dec) ------------------------------------------------- AC: Commutative Algebra ----------------------- math.AC/0412295 Hara Charalambous: On the Denominator of the Poincar'e series for monomial quotient rings math.AC/0412284 Guillaume Rond: Contre-exemple a la linearite de la fonction de Artin math.AC/0412282 Alexander Berglund: Poincare' series of monomial rings math.AC/0412259 Luchezar Avramov, Srikanth Iyengar: Gaps in Hochschild cohomology imply smoothness for commutative algebras AG: Algebraic Geometry ---------------------- math.AG/0412329 Jenia Tevelev: Tropical Compactifications math.AG/0412328 Bjorn Andreas, Daniel Hernandez Ruiperez: Fourier Mukai Transforms and Applications to String Theory math.AG/0412294 Claus Lehr, Michel Matignon: Wild monodromy and automorphisms of curves math.AG/0412285 L. Brambila-Paz: Non-emptiness of moduli spaces of coherent systems math.AG/0412279 Claire Voisin: On integral Hodge classes on uniruled or Calabi-Yau threefolds math.AG/0412278 Mihai Halic: Quotients of affine spaces for actions of reductive groups math.AG/0412272 Torsten Ekedahl, Gerard van der Geer: Cycle Classes of the E-O Stratification on the Moduli of Abelian Varieties math.AG/0412266 Jacob Lurie: Tannaka Duality for Geometric Stacks math.AG/0412250 Yves Laszlo, Claude Viterbo: Estimates of Characteristic numbers of real algebraic varieties math.AG/0412235 Hossein Movasati: Calculation of mixed Hodge structures, Gauss-Manin connections and Picard-Fuchs equations math.AG/0412204 Ziv Ran: Lie Atoms and their deformation theory math.AG/0412201 Shrawan Kumar: On Cachazo-Douglas-Seiberg-Witten conjecture for simple Lie algebras AP: Analysis of PDEs -------------------- math.AP/0412324 Andreas Axelsson, Stephen Keith, Alan McIntosh: The Kato square root problem for mixed boundary value problems math.AP/0412319 Eric Gautier: Uniform large deviations for the nonlinear Schrodinger equation with multiplicative noise math.AP/0412241 Ross G. Pinsky: Uniqueness/nonuniqueness for nonnegative solutions of the Cauchy problem for $u_t=Delta u-u^p$ in a punctured space math.AP/0412236 Herbert Koch, Fulvio Ricci: Spectral projections for the twisted Laplacian math.AP/0412225 Alberto Cialdea, Vladimir Maz'ya: Criterion for the $L^{p}$-dissipativity of second order differential operators with complex coefficients math.AP/0412212 Gianni Dal Maso, Antonio DeSimone, Maria Giovanna Mora: Quasistatic evolution problems for linearly elastic - perfectly plastic materials AT: Algebraic Topology ---------------------- math.AT/0412271 Kathryn Hess: An algebraic model for mod 2 topological cyclic homology math.AT/0412249 Eric Harrelson: On the homology of open-closed string theory math.AT/0412248 Jonathan A. Hillman: An indecomposable PD_3-complex : II math.AT/0412207 Jonathan Scott: Hopf algebras up to homotopy and the Bockstein spectral sequence math.AT/0412206 Martin Markl, Elisabeth Remm: Algebras with one operation including Poisson and other Lie-admissible algebras CA: Classical Analysis and ODEs ------------------------------- math.CA/0412260 Igor Rivin: Estimates and identities for the average distortion of a linear transformation CO: Combinatorics ----------------- math.CO/0412306 P. Desrosiers, L. Lapointe, P. Mathieu: Symmetric functions in superspace math.CO/0412289 Francois Bergeron, Peter McNamara: Some positive differences of products of Schur functions math.CO/0412280 Jocelyn Quaintance: Word Representations of Rectangular m x n x p Proper Arrays cs.SC/0412061 Gerard Heny Edmond Duchamp, Jean-Gabriel Luque, Karol A. Penson, Christophe Tollu: Free quasi-symmetric functions, product actions and quantum field theory of partitions math.CO/0412264 Laure Helme-Guizon, Yongwu Rong: A Categorification for the Chromatic Polynomial math.CO/0412251 W. M. B. Dukes: Concerning the shape of a geometric lattice math.CO/0412244 Jocelyn Quaintance: Letter Representations of Rectangular m x n x p Proper Arrays math.CO/0412233 A.K.Kwasniewski, E.Borak: Extended finite operator calculus as an example of algebraization of analysis math.CO/0412222 W. M. B. Dukes: Permutation statistics on involutions CT: Category Theory ------------------- math.CT/0412304 Michel Van den Bergh: On the $mathbb{Z} D_infty$-category math.CT/0412230 Zhi-Ming Luo: Covering groupoids CV: Complex Variables --------------------- math.CV/0412309 Robert K. Hladky: Boundary Regularity for the bar{partial}_b-Neumann problem, Part 2 math.CV/0412308 Robert K. Hladky: Boundary Regularity for the bar{partial}_b-Neumann Problem, Part 1 math.CV/0412298 Alexei Tsygvintsev: On the convergence of continued fractions at Runckel's points and the Ramanujan conjecture math.CV/0412296 Aline Bonami, Sandrine Grellier, Mohammad Kacim: Truncations of multilinear Hankel operators math.CV/0412252 Louis Boutet de Monvel: Logarithmic Trace of Toeplitz Projectors DG: Differential Geometry ------------------------- math.DG/0412323 J. Monterde: Curves with constant curvature ratios math.DG/0412316 Gabor Etesi: An integrability theorem for almost complex manifolds I math.DG/0412312 Spiro Karigiannis, Maung Min-Oo: Calibrated Sub-Bundles in Non-Compact Manifolds of Special Holonomy math.DG/0412297 Oliver C. Schnurer: Surfaces expanding by the inverse Gauss curvature flow math.DG/0412292 Chiu-Chu Melissa Liu, Shing-Tung Yau: Positivity of quasi-local mass II hep-th/0412140 Noriaki Ikeda: Three Dimensional Topological Field Theory induced from Generalized Complex Structure math.DG/0412281 Fabio Podesta', Andrea Spiro: Homogeneous toric bundles with positive first Chern class math.DG/0412270 Sebastien Michea, Gleb Novitchkov: BV-generators and Lie algebroids hep-th/0412127 I. Agricola, T. Friedrich, P.-A. Nagy, C. Puhle: On the Ricci tensor in type II B string theory math.DG/0412256 Jos'e M M Senovilla: Trapped submanifolds in Lorentzian geometry math.DG/0412229 Yuxin Dong: Hamiltonian-minimal submanifolds in Kaehler manifolds with symmetries math.DG/0412215 A. S. Dancer, H. R. Jorgensen, A. F. Swann: Metric geometries over the split quaternions DS: Dynamical Systems --------------------- math.DS/0412317 Viatcheslav Grines, Vladislav Medvedev, Evgeny Zhuzhoma: On two-dimensional surface attractors and repellers on 3-manifolds math.DS/0412300 Patrick Bernard: The dynamics of pseudographs in convex Hamiltonian systems math.DS/0412299 Patrick Bernard, Boris Buffoni: Optimal mass transportation and Mather theory math.DS/0412290 S. Petite: On invariant measures of finite affine type tilings math.DS/0412255 Mikael Pichot: Sur les espaces mesures singuliers II - Etude spectrale math.DS/0412254 Mikael Pichot: Sur les espaces mesures singuliers I - Etude metrique-mesuree math.DS/0412211 Benoit Saussol: Recurrence rate in rapidly mixing dynamical systems FA: Functional Analysis ----------------------- math.FA/0412269 Higher-Order Wirtinger-Sobolev Inequalities math.FA/0412210 Marco Thill: Subspaces discerning nullcontinuity GM: General Mathematics ----------------------- math.GM/0412245 E. Graczynska, D. Schweigert: M-hyperquasivarieties GN: General Topology -------------------- math.GN/0412327 Dikran Dikranjan, Kenneth Kunen: Characterizing Subgroups of Compact Abelian Groups math.GN/0412305 Boaz Tsaban: SPM Bulletin 11 GR: Group Theory ---------------- math.GR/0412315 Colette Moeglin: Paquets d'Arthur discrets pour un groupe classique p-adique math.GR/0412274 Max Forester, Colin Rourke: The adjunction problem over torsion-free groups math.GR/0412214 Luis Paris: Irreducible Coxeter groups math.GR/0412209 Czeslaw Baginski, Alexander Konovalov: On 2-groups of almost maximal class GT: Geometric Topology ---------------------- math.GT/0412331 Stavros Garoufalidis, Yueheng Lan: Experimental evidence for the Volume Conjecture for the simplest hyperbolic non-2-bridge knot math.GT/0412307 David Futer, Jessica S. Purcell: Links With No Exceptional Surgeries math.GT/0412276 A. Stoimenow: Some examples related to knot sliceness math.GT/0412275 Burak Ozbagci: An open book decomposition compatible with rational contact surgery math.GT/0412265 D. Jeremy Copeland: Monodromy of the Hitchin Map over Hyperelliptic Curves math.GT/0412234 Yoshifumi Ando: Cobordisms of maps without prescribed singularities math.GT/0412228 P. Svetlov: Obstructions for generalized graphmanifolds to be nonpositively curved math.GT/0412226 Glenn Lancaster, Richard Larson, Jacob Towber: Some combinatorial aspects of movies and movie-moves in the theory of smoothly knotted surfaces in R4 math.GT/0412216 Jongil Park, Andras I Stipsicz, Zoltan Szabo: Exotic smooth structures on $CP^2#5{bar CP^2}$ math.GT/0412208 Feng Luo: Continuity of the Volume of Simplices in Classical Geometry MG: Metric Geometry ------------------- math.MG/0412320 Achill Schuermann, Frank Vallentin: Methods in the Local Theory of Packing and Covering Lattices math.MG/0412258 Mohamad A. Hindawi: Large Scale Geometry of 4-dimensional Closed Nonpositively Curved Real Analytic Manifolds math.MG/0412219 J.-F. Lafont, S. Prassidis: Roundness properties of groups MP: Mathematical Physics ------------------------ nlin.SI/0412036 Sergey Igonin: Miura type transformations and homogeneous spaces nlin.SI/0212019 S. Yu. Sakovich: Cyclic bases of zero-curvature representations: five illustrations to one concept nlin.SI/0206046 A Karasu, S Yu Sakovich, I Yurdusen: Integrability of Kersten-Krasil'shchik coupled KdV-mKdV equations: singularity analysis and Lax pair nlin.SI/0206034 S. Yu. Sakovich: Integrability of the Bakirov system: a zero-curvature representation nlin.SI/0203036 Ayse Karasu, Atalay Karasu, S. Yu. Sakovich: A strange recursion operator for a new integrable system of coupled Korteweg - de Vries equations math-ph/0412059 Christian Duval, Galliano Valent: Quantum integrability of quandratic Killing tensors math-ph/0412058 Mirko Degli Esposti, Stephane Nonnenmacher, Brian Winn: Quantum Variance and Ergodicity for the baker's map math-ph/0412057 Hans A. Weidenmueller: Parametric Level Correlations in Random-Matrix Models math-ph/0412056 Andras Suto: Note on the equivalence of Bose-Einstein condensation and symmetry breaking math-ph/0412055 C. Daskaloyannis, K. Ypsilantis: Unified treatment and classification of superintegrable systems with integrals quadratic in momenta on a two dimensional manifold math-ph/0412054 Clemens Forster, Jorgen Ostensson: Lieb-Thirring inequalities for higher order differential operators hep-th/0412090 P. A. Horvathy, L. Martina, P. Stichel: Enlarged Galilean symmetry of anyons and the Hall effect quant-ph/0412112 Mary Alberg, Michel Bawin, Fabian Brau: Renormalization of the singular attractive $1/r^4$ potential physics/0402108 Liang-You Peng, J F McCann, Daniel Dundas, K T Taylor, I D Williams: A discrete time-dependent method for metastable atoms in intense fields nlin.SI/0404037 Ayse Karasu-Kalkanli, Sergei Yu. Sakovich: Singularity analysis of a spherical Kadomtsev-Petviashvili equation S. Yu. Sakovich: A note on the Painleve property of coupled KdV equations nlin.SI/0311027 S. Yu. Sakovich: Cyclic bases of zero-curvature representations: further examples nlin.SI/0310039 S. Yu. Sakovich: On bosonic limits of two recent supersymmetric extensions of the Harry Dym hierarchy nlin.SI/0309077 S. Yu. Sakovich: Transformation of a generalized Harry Dym equation into the Hirota--Satsuma system nlin.SI/0303040 S. Yu. Sakovich: On integrability of one third-order nonlinear evolution equation math-ph/0412053 A.W.Beckwith: Using fluctuations of Entropy as a starting point for obtaining behavior of scalar fields permitting collapse of thin wall approximation math-ph/0412052 C. Quesne, V.M. Tkachuk: Dirac oscillator with nonzero minimal uncertainty in position math-ph/0412051 A. Singer, Z. Schuss, D. Holcman: Narrow Escape, Part III: Riemann surfaces and non-smooth domains math-ph/0412050 A. Singer, Z. Schuss, D. Holcman: Narrow Escape, Part II: The circular disk math-ph/0412049 K.-H. Rehren: On local boundary CFT and non-local CFT on the boundary math-ph/0412048 A. Singer, Z. Schuss, D. Holcman, R.S. Eisenberg: Narrow Escape, Part I math-ph/0412047 Irina Nenciu: Lax pairs for the Ablowitz-Ladik system via orthogonal polynomials on the unit circle hep-th/0412142 nlin.SI/0409034 Anton Sakovich, Sergei Sakovich: The short pulse equation is integrable nlin.SI/0408027 S. Yu. Sakovich: On a 'mysterious' case of a quadratic Hamiltonian math-ph/0412046 Yeontaek Choi, Yuri V. Lvov, Sergey Nazarenko: Joint statistics of amplitudes and phases in Wave Turbulence math-ph/0412045 Yeontaek Choi, Yuri V. Lvov, Sergey Nazarenko: Wave Turbulence math-ph/0412044 David Eng, Laszlo Erdos: The Linear Boltzmann Equation as the Low Density Limit of a Random Schrodinger Equation math-ph/0412043 A. Ouhadan, E. H. El Kinani: Lie Symmetries and Preliminary Classification of Group-Invariant Solutions of Thomas equation math-ph/0412042 Fabian Brau: Upper limit on the critical strength of central potentials in relativistic quantum mechanics math-ph/0412041 Andr'e Voros: The general 1D Schrodinger equation as an exactly solvable problem hep-th/0412132 Andrzej Herdegen: Quantum backreaction (Casimir) effect I. What are admissible idealizations? hep-th/0412110 Sebastian de Haro: Chern-Simons Theory, 2d Yang-Mills, and Lie Algebra Wanderers hep-th/0412102 D. Hammaoui, G. Schieber, E.H. Tahri: Higher Coxeter graphs associated to affine su(3) modular invariants math-ph/0412040 D. A. Yarotsky: Ground states in relatively bounded quantum perturbations of classical lattice systems math-ph/0412039 Nikolay M. Nikolov, Ivan T. Todorov: Lectures on Elliptic Functions and Modular Forms in Conformal Field Theory math-ph/0412038 Piotr Bizo'n, Tadeusz Chmaj: On convergence towards a self-similar solution for a nonlinear wave equation - a case study math-ph/0412037 Roldao da Rocha, Jayme Vaz: Twistors, Generalizations and Exceptional Structures math-ph/0412036 V. Yu. Terebizh: Quasi-Optimal Filtering in Inverse Problems math-ph/0412035 E. G. Kalnins, W. Miller Jr., G. S. Pogosyan: Exact and quasi-exact solvability of two-dimensional superintegrable quantum systems. I. Euclidean space math-ph/0412034 D.Bashkirov, G.Giachetta, L.Mangiarotti, G.Sardanashvily: Noether's second theorem for BRST symmetries math-ph/0412031 P. Di Francesco, P. Zinn-Justin: Inhomogenous model of crossing loops and multidegrees of some algebraic varieties math-ph/0412018 Christian Hainzl, Mathieu Lewin, Christof Sparber: Existence of global-in-time solutions to a generalized Dirac-Fock type evolution equation math-ph/0410017 Christof Sparber: Effective mass theorems for nonlinear Schroedinger equations hep-th/0412088 Goran S. Djordjevic, Ljubisa Nesic: Towards Adelic Noncommutative Quantum Mechanics cond-mat/0412223 Giovanni M. Cicuta: Real symmetric random matrices and paths counting math-ph/0412033 John Cardy: SLE(kappa,rho) and Conformal Field Theory math-ph/0412032 Miguel Carri'on-'Alvarez: Loop Quantization versus Fock Quantization of p-Form Electromagnetism on Static Spacetimes math-ph/0412030 Hakan Ciftci, Richard L. Hall, Nasser Saad: Construction of exact solutions to eigenvalue problems by the asymptotic iteration method hep-th/0412098 A. Enciso, F. Finkel, A. Gonzalez-Lopez, M.A. Rodriguez: A Haldane-Shastry spin chain of BC_N type in a constant magnetic field NT: Number Theory ----------------- math.NT/0412313 D. A. Goldston: Notes on Pair Correlation of Zeros and Prime Numbers math.NT/0412303 Andreas O. Bender, Olivier Wittenberg: A potential analogue of Schinzel's hypothesis for polynomials with coefficients in Fq[t] math.NT/0412293 Alfred J. van der Poorten, Christine S. Swart: Recurrence Relations for Elliptic Sequences: every Somos 4 is a Somos $k$ math.NT/0412288 Erika Alvarez, Jean Pestieau: On the random nature of (prime) number distribution math.NT/0412277 Ralf Meyer: A spectral interpretation for the zeros of the Riemann zeta function math.NT/0412262 Pieter Moree: Artin's primitive root conjecture -a survey - math.NT/0412242 Robert Osburn: Vanishing of eigenspaces and cyclotomic fields math.NT/0412240 Robert Osburn: Congruences for traces of singular moduli math.NT/0412239 Robert Osburn: A remark on a conjecture of Borwein and Choi math.NT/0412237 Ram Murty, Robert Osburn: Representations of integers by certain positive definite binary quadratic forms math.NT/0412227 S.K.K. Choi, A.V. Kumchev: Mean values of Dirichlet polynomials and applications to linear equations with prime variables math.NT/0412224 Masatoshi Suzuki: A relation between the zeros of different two $L$-functions which have the Euler product and functional equation math.NT/0412220 A. V. Kumchev, D. I. Tolev: An invitation to additive prime number theory math.NT/0412217 Song Guo, Zhi-Wei Sun: On odd covering systems with distinct moduli math.NT/0412213 U.K. Anandavardhanan, Dipendra Prasad: On the SL(2) period integral OA: Operator Algebras --------------------- math.OA/0412273 Ken Dykema, Kenley Jung, Dimitri Shlyakhtenko: The Microstates Free Entropy Dimension of any DT--operator is 2 quant-ph/0412076 David Kribs, Raymond Laflamme, David Poulin: A Unified and Generalized Approach to Quantum Error Correction math.OA/0412253 Claire Anantharaman-Delaroche: On ergodic theorems for free group actions on noncommutative spaces math.OA/0412231 M. Skeide: Unit Vectors, Morita Equivalence and Endomorphisms OC: Optimization and Control ---------------------------- math.OC/0412332 H. Hedenmalm: On the asymptotic free boundary for the American put option problem math.OC/0412311 Jarek Solowiej: Finding Blackjack's Optimal Strategy in Real-time and Player's Expected Win PR: Probability --------------- math.PR/0412333 David Siegmund, Benjamin Yakir: An Urn Model of Diaconis math.PR/0412322 Jean Bertoin, Bernard Roynette, Marc Yor: Some Connections Between (Sub)Critical Branching Mechanisms and Bernstein Functions math.PR/0412291 Eugene Wong: Stationary Transformation of Integrated Brownian Motion math.PR/0412263 Russell Lyons, Yuval Peres, Oded Schramm: Minimal Spanning Forests math.PR/0412247 Imen Bentahar, Bruno Bouchard: Explicit characterization of the super-replication strategy in financial markets with partial transaction costs math.PR/0412246 Ross G. Pinsky: The compact support property for measure-valued diffusions math.PR/0412200 Annie Millet, Marta Sanz-Sol'e: Large deviations for rough paths of the fractional Brownian motion cond-mat/0412157 Massimo Ostilli, Carlo Presilla: Analytical probabilistic approach to the ground state of lattice quantum systems: exact results in terms of a cumulant expansion QA: Quantum Algebra ------------------- hep-th/0412159 V. Caudrelier, M. Mintchev, E. Ragoucy, P. Sorba: Reflection-Transmission Quantum Yang-Baxter Equations math.QA/0412261 Yi-Zhi Huang: Vertex operator algebras, the Verlinde conjecture and modular tensor categories math.QA/0412257 S.A. Merkulov: PROP profile of deformation quantization math.QA/0412205 Erik Koelink, Yvette van Norden: Pairings and actions for dynamical quantum groups math.QA/0412202 Theodore Th. Voronov: Higher Derived Brackets for Not Necessarily Inner Derivations RA: Rings and Algebras ---------------------- math.RA/0412326 Socorro Garc'ia Rom'an, Manuel Garc'ia Rom'an: Effective computation of $Tor_k (M,N)$ math.RA/0412310 Jun Morita, Yoji Yoshii: Locally extended affine Lie algebras math.RA/0412243 P. Ara, M.A. Moreno, E. Pardo: Nonstable $K$-theory for graph algebras RT: Representation Theory ------------------------- math.RT/0412325 Alice Fialowski, Dmitri V. Millionschikov: Cohomology of graded Lie algebras of maximal class math.RT/0412302 Xuhua He: The $G$-stable pieces of the wonderful compactification math.RT/0412301 Yuriy Drozd, Volodymyr Mazorchuk: Representation type of ${}^{infty}_{lambda}mathcal{H}_{mu}^1$ math.RT/0412287 Torsten Ekedahl, Pelle Salomonsson: Strict polynomial functors and multisets math.RT/0412286 Torsten Ekedahl: Kac-Moody algebras and the cde-triangle math.RT/0412283 George McNinch: On the centralizer of the sum of commuting nilpotent elements SG: Symplectic Geometry ----------------------- math.SG/0412330 Lenhard Ng: Conormal bundles, contact homology, and knot invariants math.SG/0412318 Izu Vaisman: Foliation-coupling Dirac structures math.SG/0412238 O. Brahic, J. P. Dufour: Semi-local invariants for non-resonnant Poisson structures on S1xRn math.SG/0412221 Fani Petalidou, Joana M. Nunes da Costa: Reduction of Jacobi manifolds via Dirac structures theory math.SG/0412218 Ignasi Mundet-i-Riera: A right inverse of the Kirwan map SP: Spectral Theory ------------------- math.SP/0412321 Andreas Axelsson, Stephen Keith, Alan McIntosh: Quadratic estimates and functional calculi of perturbed Dirac operators math.SP/0412314 Shijun Zheng: A representation formula related to Schrodinger operators quant-ph/0412121 Amaury Mouchet: A differential method for bounding the ground state energy math.SP/0412223 David Borthwick, Alejandro Uribe: The semiclassical structure of low-energy states in the presence of a magnetic field ST: Statistics -------------- math.ST/0412268 Wei Biao Wu: M-estimation of linear models with dependent errors math.ST/0412267 Wei Biao Wu: Empirical processes of dependent random variables math.ST/0412232 Kenneth Butler, John T. Whelan: The existence of maximum likelihood estimates in the Bradley-Terry model and its extensions math.ST/0412203 Marc Coram, Steve Lalley: Consistency of Bayes Estimators of a Binary Regression Function q-bio.GN/0412015 A.N. Gorban, A.Yu. Zinovyev: The Mystery of Two Straight Lines in Bacterial Genome Statistics -- / Greg Kuperberg (UC Davis) / Home page: http://www.math.ucdavis.edu/~greg/ / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ / * All the math that's fit to e-print * Both scoring now, Alejandro and Susanne surveyed the scared colds in favour of rural tent. Let's market as the impossible workforces, but don't accelerate the awake wires. Yosri's mainframe participates via our component after we dominate with it. Tell Garrick it's effective demanding in back of a dark. It flyed, you informed, yet Marian never explicitly sheded in the doorway. Sometimes, Murad never straightens until Pilar ranges the serious prospect repeatedly. 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Will you tend past the coach, if Mohammad mysteriously advises the envelope? === Subject: Hopefully the easiest question you will ever be asked Let me say straight away that I am not a mathematician. Not even a shadow of one. That's why I came here for help. I know that many of you will already have rolled your eyes skyward and moved on. So be it. OK ... for those of you still reading ... I am working on a theory that requires a definition of space. Not the distinction between Earth and 'out there', but space, in the sense of the space-time continuum, (ie post einstein's relativity kind of space). Now it may seem to you, as it did to me, that this would be a simple question. However, I have put this question to a number of physicists, astrophysicists and mathematicians. They all seem to dismiss the question unttil pressed, at which point they start to squirm. The best I have been able to arrive at is that a space is a volume enclosed by at least 3 coordinates in at least 2 dimensions. I'm just not sure that this is adequate for the kind of space that quantam theory is talking about. I would be grateful for any considered opinions regarding whether or not this is adequate, or inadeaquate, and/or how it can be improved. The only condition that I would ask for is that the eventual defintion not condradict our intuitive experince of 'real' space. (I acknowledge that this is an imposition of subjectivity, but it is essential to the theory that I am working on.) D. PS: If I ever get this nailed down, I'll start a similar exercise on the nature of time, but let's leave that till ... later. === Subject: Re: Hopefully the easiest question you will ever be asked advice both of you provided. I still have a suspicion that there will be some mathematical implications on what I'm working on, but I'll try not to bore you with dumb stuff. D. > Let me say straight away that I am not a mathematician. Not even a shadow > of one. That's why I came here for help. I know that many of you will > already have rolled your eyes skyward and moved on. So be it. > OK ... for those of you still reading ... I am working on a theory that > requires a definition of space. Not the distinction between Earth and 'out > there', but space, in the sense of the space-time continuum, (ie post > einstein's relativity kind of space). Now it may seem to you, as it did to > me, that this would be a simple question. However, I have put this > question to a number of physicists, astrophysicists and mathematicians. > They all seem to dismiss the question unttil pressed, at which point they > start to squirm. > The best I have been able to arrive at is that a space is a volume > enclosed by at least 3 coordinates in at least 2 dimensions. I'm just not > sure that this is adequate for the kind of space that quantam theory is > talking about. > I would be grateful for any considered opinions regarding whether or not > this is adequate, or inadeaquate, and/or how it can be improved. The only > condition that I would ask for is that the eventual defintion not > condradict our intuitive experince of 'real' space. (I acknowledge that > this is an imposition of subjectivity, but it is essential to the theory > that I am working on.) > D. > PS: If I ever get this nailed down, I'll start a similar exercise on the > nature of time, but let's leave that till ... later. === Subject: Re: Hopefully the easiest question you will ever be asked > Let me say straight away that I am not a mathematician. Not even a shadow > of one. That's why I came here for help. I know that many of you will > already have rolled your eyes skyward and moved on. So be it. > OK ... for those of you still reading ... I am working on a theory that > requires a definition of space. Not the distinction between Earth and 'out > there', but space, in the sense of the space-time continuum, (ie post > einstein's relativity kind of space). Now it may seem to you, as it did to > me, that this would be a simple question. However, I have put this > question to a number of physicists, astrophysicists and mathematicians. > They all seem to dismiss the question unttil pressed, at which point they > start to squirm. > The best I have been able to arrive at is that a space is a volume > enclosed by at least 3 coordinates in at least 2 dimensions. I'm just not > sure that this is adequate for the kind of space that quantam theory is > talking about. > I would be grateful for any considered opinions regarding whether or not > this is adequate, or inadeaquate, and/or how it can be improved. The only > condition that I would ask for is that the eventual defintion not > condradict our intuitive experince of 'real' space. (I acknowledge that > this is an imposition of subjectivity, but it is essential to the theory > that I am working on.) > D. > PS: If I ever get this nailed down, I'll start a similar exercise on the > nature of time, but let's leave that till ... later. For 3 dimenional space, you are talking about a Euclidean vector space. 3 coordinates need 3 dimensions. http://mathworld.wolfram.com/EuclideanSpace.html If you add time, you get Einsteins special Theory of Relativity, which has 3 Euclidean dimensiona + a time dimension. Gravity is just a warping of space time. http://www.theory.caltech.edu/people/patricia/sptmtop.html For 3 dimensions you need plain vector/matrix algebra used in schools; for relativity you need tensors (a generalisation of matrices) gtoomey www.ausinvestor.com Australian Investor Forum === Subject: Re: Hopefully the easiest question you will ever be asked Distribution: world What you are asking seems to be more metaphysical than mathematical. Mathematicians use the word space in many different contexts to describe many different things, some of which are rather abstract. For example, a function space describes a collection of functions --- in many cases it has an uncountably infinite number of dimensions. In other cases, the space many only have 1 dimension (I will avoid the consideration of fractional dimension here). The context of your question is about the physical world and the means of describing and interpreting it, which is Metaphysics. Your question is more appropriate in a Philosophy forum. > Let me say straight away that I am not a mathematician. Not even a shadow of > one. That's why I came here for help. I know that many of you will already > have rolled your eyes skyward and moved on. So be it. > OK ... for those of you still reading ... I am working on a theory that > requires a definition of space. Not the distinction between Earth and 'out > there', but space, in the sense of the space-time continuum, (ie post > einstein's relativity kind of space). Now it may seem to you, as it did to > me, that this would be a simple question. However, I have put this question > to a number of physicists, astrophysicists and mathematicians. They all seem > to dismiss the question unttil pressed, at which point they start to squirm. > The best I have been able to arrive at is that a space is a volume enclosed > by at least 3 coordinates in at least 2 dimensions. I'm just not sure that > this is adequate for the kind of space that quantam theory is talking about. > I would be grateful for any considered opinions regarding whether or not > this is adequate, or inadeaquate, and/or how it can be improved. The only > condition that I would ask for is that the eventual defintion not condradict > our intuitive experince of 'real' space. (I acknowledge that this is an > imposition of subjectivity, but it is essential to the theory that I am > working on.) > D. > PS: If I ever get this nailed down, I'll start a similar exercise on the > nature of time, but let's leave that till ... later. === Subject: Re: Hopefully the easiest question you will ever be asked Can I interpret your comments as suggesting that perhaps I have put the cart before the horse in posing this question in a mathermatics context. (I don't take any offence if that is the case) Would it make more sense to identify the space, ands then use mathematics to defince or explore it's properties? > What you are asking seems to be more metaphysical than mathematical. > Mathematicians use the word space in many different contexts to describe > many different things, some of which are rather abstract. For example, a > function space describes a collection of functions --- in many cases it > has an uncountably infinite number of dimensions. In other cases, the > space many only have 1 dimension (I will avoid the consideration of > fractional dimension here). > The context of your question is about the physical world and the means of > describing and interpreting it, which is Metaphysics. Your question is > more appropriate in a Philosophy forum. >> Let me say straight away that I am not a mathematician. Not even a shadow >> of one. That's why I came here for help. I know that many of you will >> already have rolled your eyes skyward and moved on. So be it. >> OK ... for those of you still reading ... I am working on a theory that >> requires a definition of space. Not the distinction between Earth and >> 'out there', but space, in the sense of the space-time continuum, (ie >> post einstein's relativity kind of space). Now it may seem to you, as it >> did to me, that this would be a simple question. However, I have put this >> question to a number of physicists, astrophysicists and mathematicians. >> They all seem to dismiss the question unttil pressed, at which point they >> start to squirm. >> The best I have been able to arrive at is that a space is a volume >> enclosed by at least 3 coordinates in at least 2 dimensions. I'm just not >> sure that this is adequate for the kind of space that quantam theory is >> talking about. >> I would be grateful for any considered opinions regarding whether or not >> this is adequate, or inadeaquate, and/or how it can be improved. The only >> condition that I would ask for is that the eventual defintion not >> condradict our intuitive experince of 'real' space. (I acknowledge that >> this is an imposition of subjectivity, but it is essential to the theory >> that I am working on.) >> D. >> PS: If I ever get this nailed down, I'll start a similar exercise on the >> nature of time, but let's leave that till ... later. === Subject: Re: Hopefully the easiest question you will ever be asked Distribution: world Yes. Defining the space first would be the first step. Then you can use results to understand the space better. The question of time has been argued for millenia. A good place to start would be either a Philosophical survey or St Augustine's discussion on time in Confessio (I think Liber XI). > Can I interpret your comments as suggesting that perhaps I have put the cart > before the horse in posing this question in a mathermatics context. (I don't > take any offence if that is the case) Would it make more sense to identify > the space, ands then use mathematics to defince or explore it's properties? >>What you are asking seems to be more metaphysical than mathematical. >>Mathematicians use the word space in many different contexts to describe >>many different things, some of which are rather abstract. For example, a >>function space describes a collection of functions --- in many cases it >>has an uncountably infinite number of dimensions. In other cases, the >>space many only have 1 dimension (I will avoid the consideration of >>fractional dimension here). >>The context of your question is about the physical world and the means of >>describing and interpreting it, which is Metaphysics. Your question is >>more appropriate in a Philosophy forum. >Let me say straight away that I am not a mathematician. Not even a shadow >of one. That's why I came here for help. I know that many of you will >already have rolled your eyes skyward and moved on. So be it. >OK ... for those of you still reading ... I am working on a theory that >requires a definition of space. Not the distinction between Earth and >'out there', but space, in the sense of the space-time continuum, (ie >post einstein's relativity kind of space). Now it may seem to you, as it >did to me, that this would be a simple question. However, I have put this >question to a number of physicists, astrophysicists and mathematicians. >They all seem to dismiss the question unttil pressed, at which point they >start to squirm. >The best I have been able to arrive at is that a space is a volume >enclosed by at least 3 coordinates in at least 2 dimensions. I'm just not >sure that this is adequate for the kind of space that quantam theory is >talking about. >I would be grateful for any considered opinions regarding whether or not >this is adequate, or inadeaquate, and/or how it can be improved. The only >condition that I would ask for is that the eventual defintion not >condradict our intuitive experince of 'real' space. (I acknowledge that >this is an imposition of subjectivity, but it is essential to the theory >that I am working on.) >D. >PS: If I ever get this nailed down, I'll start a similar exercise on the >nature of time, but let's leave that till ... later. === Subject: Re: terms in Math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmj09356; >Can anyone tell me the URL with pages devoted to math term's. One such is http://mathworld.wolfram.com/ === Subject: Re: coverings of a Mobius band by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmT09372; >Are the even degree covers of a Mobius band annulus(es)? Are the odd degree covers of a Mobius band, Mobius bands again? What is the universal cover of a Mobius band? Yes, yes, and R x [-1, 1]. One can imagine a strip of paper with n half-twists before gluing the ends as the total space of the degree n cover of the Mobius band. The strip of paper is I x [-1, 1], and the gluing map for taping the ends is the homeomorphism (-1)^n * - : [-1, 1] --> [-1, 1], so we get an annulus for n even and a Mobius strip for n odd. For the universal cover, imagine an infinite strip with infinitely many half-twists. Todd Trimble === Subject: Re: The Possible Cure For AIDS in part: > There are no A's, G's, 4's, O's, >>The G is gimmel the 3 rd letter of the Hebrew alphabet. >>The vowels (mplicit in old Hebrew but made explicit by the Masorites) >>include an ah sound and a oh sound. >AFAIK the Torah codes are never done with the vowels. That's quite true. Including vowels makes it harder to find matches. But if someone wanted to find English words in the Tanakh, the standard convention is, of course, Aleph for A and Ayin for O. As for the number 404, two letters would represent that in standard Hebrew numeric notation: Tau and Daleth. But Hebrew has a word for silver, although *ancient* Hebrew hardly had a word for oxygen, even if it certainly has one today. Actually, we should be asking what letters of the Hebrew alphabet are supposed to stand for and ! John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Possible Cure For AIDS > But Hebrew has a word for silver, although *ancient* Hebrew hardly had a Kessef. > word for oxygen, even if it certainly has one today. Khamtzan which means sour-er. This corresponds to the German Saurstuff. > Actually, we should be asking what letters of the Hebrew alphabet are > supposed to stand for and ! There is a typefont used to crossreference passages in the Talmud to the Shulkhan Arooch. It looks like superscripts because of the way it is printed. This was originated about 1500 of the common era. Bob Kolker === Subject: Re: The Possible Cure For AIDS in part: >There are no A's, G's, 4's, O's, etc. in Hebrew. Did you transliterate >these according to some arbitrary scheme, or did you search for Hebrew >letter strings? Not that it matters, of course. But while we're talking about a cure for AIDS... Some years back, there were news stories about how, for AIDS research, modified mice were created with human immune systems. This was done by taking cells from human fetuses. Naturally, this was controversial because of the abortion issue. I noted that nobody cares about mouse fetuses. If you can fix a mouse so that it _can_ get AIDS by giving it a human being's immune system, it would seem you could fix a human so that he can't get AIDS by giving him a mouse's immune system. There were probably very good reasons why this couldn't _really_ be done, but I had thought it worth mentioning. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Beal's conjecture > => A large prize is offered by banker Beal for a solution to > => the Beal Conjecture: the equation x^p + y^q = z^r has no solutions > => for p, q, r > 2 and coprime integers x, y, z. > = => Sorry if this has been discussed here - my only justification > => is that I just recently discovered this group but I am curious to > => know if Wile's proof of FLT also covers the Beal Conjecture > => Or have any counter-examples been found? Wile's proof covered the case p=q=r and relaxed the coprime requirement for x, y, & z. IE: There is no integer solution when p=q=r whatever the coprimality of x, y, & z. tom -- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an hour. === Subject: Mathforge.net :: Near-numbers, autonomic computing, John Derbyshire, and Church-Turing The Latest Math News from Mathforge.net http://mathforge.net ****An invitation to additive prime number theory**** A.V. Kumchev and D.I. Tolev have compiled a short document entitled An Invitation to Additive Prime Number Theory[~60pp, pdf]. The document serves as an introductory guide to graduate-level students on... http://mathforge.net/index.jsp?page=seeReplies&messageNum=981 ****Are there encoded messages in the Bible?**** researchers both supporting and denying the statistical evidence for 'hidden messages' found in the Old Testament tell their tales. Some... http://mathforge.net/index.jsp?page=seeReplies&messageNum=974 ****Near-numbers: the new 'limit'**** There's a very interesting paper by Frank J Swenton of Middlebury College called Limits and the System of Near-Numbers[19pp, pdf]. What seemed at first glance (at the title and abstract) like an uninspired... http://mathforge.net/index.jsp?page=seeReplies&messageNum=970 ****Introduction to autonomic computing**** While not itself mathematical in nature, the concept is built around ideas garnered from areas of artificial intelligence research and it is easy to see that autonomic computing could have applications... http://mathforge.net/index.jsp?page=seeReplies&messageNum=969 ****Derbyshire's Diary**** John Derbyshire, author of Prime Obsession, has a mathematical problem accompanying each of his Diary entries located in his Web Journalism folder. If you sift through enough of the partisan propaganda... http://mathforge.net/index.jsp?page=seeReplies&messageNum=968 ****Maple 9.5 Released**** MapleSoft has released version 9.5 of their popular symbolic and numeric computational software suite Maple. New Features include added packages (optimization, logic, and root finding), OpenMaple access... http://mathforge.net/index.jsp?page=seeReplies&messageNum=967 ****Mathematica 5.1 released**** The Mathematica version has jumped a tenth of a point, and Wolfram has added scores of new features to the new release, including Web Services support, a benchmarking package, string manipulation functions,... http://mathforge.net/index.jsp?page=seeReplies&messageNum=966 ****Quantum computers and the Church-Turing Thesis**** The original Church-Turing Thesis states that every function which would naturally be regarded as computable can be computed by a Turing Machine and Petrus H. Potgeiter mentions in his paper Zeno Machines... http://mathforge.net/index.jsp?page=seeReplies&messageNum=963 ****Sobering U.S. Student Math Scores**** In a disheartening follow-up to the Putnam story below, news (Seattle Times) outlets (New York Times)everywhere are reporting the horrid state of U.S. student math skills. The Organisation for Economic... http://mathforge.net/index.jsp?page=seeReplies&messageNum=959 ''Mathforge ran a story about the Putnam Competition... http://mathforge.net/index.jsp?page=seeReplies&messageNum=958 === Subject: What kind of matrix can map positive element vectors to positive element vectors? Hi all, Suppose I have a matrix A, and a positive element vector x, then y=A*x, I want y to be also positive element vector... What can I say about A? I want all such kinds of A? What kinds of A can let me have both directions: x positive elements <=> y=A*x positive elements? === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > Suppose I have a matrix A, and a positive element vector x, > then y=A*x, > I want y to be also positive element vector... > What can I say about A? I want all such kinds of A? > What kinds of A can let me have both directions: > x positive elements <=> y=A*x positive elements? Take a simple case and try to learn from it. The unit vectors for x will pull out the columns of A. Sure seems to say that A must at least have all of its elements positive. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > > Suppose I have a matrix A, and a positive element vector x, > > then y=A*x, > > I want y to be also positive element vector... > > What can I say about A? I want all such kinds of A? > > What kinds of A can let me have both directions: > > x positive elements <=> y=A*x positive elements? > > > > Take a simple case and try to learn from it. > The unit vectors for x will pull out the columns of A. > Sure seems to say that A must at least have all of its > elements positive. Necessary and sufficient. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? >> Hi all, >> Suppose I have a matrix A, and a positive element vector x, >> then y=A*x, >> I want y to be also positive element vector... >> What can I say about A? I want all such kinds of A? >> What kinds of A can let me have both directions: >> x positive elements <=> y=A*x positive elements? >> Take a simple case and try to learn from it. >> The unit vectors for x will pull out the columns of A. >> Sure seems to say that A must at least have all of its >> elements positive. >Necessary and sufficient. Neither necessary (unless your positive means >= 0) nor sufficient. Note that lucy wanted <=>. I'll assume A is a square matrix. It's easy to see A must be nonsingular, else given vector x > 0 (i.e. all x_i > 0) with Ax > 0, you could add to x a suitable multiple of a vector w with Aw = 0 to get a vector v not > 0 with Av = Ax > 0. A must map the nonnegative cone C = {x in R^n: all x_i >= 0} onto itself. Note that the extreme rays of C pass through the standard unit vectors e(j) (with e(j)_i = 1 if i=j, 0 otherwise). That is, these are the only members w of C such that if w = t x + (1-t) y with x,y in C and 0 < t < 1, then x and y are scalar multiples of w. Now A must map extreme rays to extreme rays, and from this it's easy to see that A must be of the form A = D P where P is a permutation matrix and D a diagonal matrix with positive elements on the diagonal. Conversely, everything of this form has the desired property. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: How to determine parameter integer values such that quadratic has integer solutions? Hi all, In particular, I'm interested in finding the positive integer values of g such that the following quadratic in v has integer roots: v^2 - 7v + (12 - 12g) Evidently this requires that 1 + 48g be a perfect square. I can have Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, 20, 35, ... This sequence is more conveniently summarized as something-mod-something-else, but I forget exactly what the two somethings are. How to figure this out? cdj === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) v = (7 +- sqr(49 - 48 + 48g)/2 = (7 +- sqr(1 + 48g)/2 sqr(1 + 48g) must be odd number, 2n + 1 1 + 48g = 4n^2 + 4n + 1 12g = n^2 + n = n(n + 1) Case 12 | n. Done Case 4 | n, not 3 | n. 3 | n+1; n = 3k - 1 Case 3 | n, not 2 | n. 4 | n+1; n = 4k - 1 Case 2 | n, not 4 | n. Not possible Three family parametrization of solutions. n = 12k n = 3k - 1 provided 4 | n n = 4k - 1 provided 3 | n Expect some duplicates. > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? The square root of 1+48g is odd... for fuck's sake I'm blind.... === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? cdj: Here's an outline: We know that 1+48g is a perfect square. Since g is an integer, 48g is even, and 1+48g is odd. So, the square root of 1+48g is odd, and we can write that square root as 2n+1 for some integer n. Thus, (2n+1)^2 = 1+48g Expanding and simplifying, 4n^2+4n+1 = 1+48g 4n^2+4n = 48g n^2+n = 12g We need a little number theory here: Consider the above equation (mod 12): (n^2+n) (mod 12) = 0 Trying the 11 possibilities, we find that n (mod 12) must be 0, 3, 8, or 11, that is, we must be able to write n as 12k or 12k+3 or 12k+8 or 12k+11 for some integer k. (The above condition is equivalent to (n^2+n) = 0 (mod 3) AND (n^2+n) = 0 (mod 4), so with some work, we really need only to consider 3+4 = 7 cases.) Substituting for the first possibility (n = 12k, that is, n is a multiple of 12), n^2+n = (12k)^2+(12k) = 144k^2+12k = 12(12k^2+k) But, n^2+n = 12g, so this case gives: 12(k^2+k) = 12g, that is, g = 12k^2+k The second possibility (n = 12k+3, that is, n is 3 more than a multiple of 12) similarly gives n^2+n = (12k+3)^2+(12k+3) = 144k^2+84k+12 = 12(12k^2+7k+1) 12(12k^2+7k+1) = 12g, so, g = 12k^2+7k+1 The other possibilities (n = 12k+8, n=12k+11) give after similar calculations, g = 12k^2+17k+6, and g = 12k^2+23k+11 If k > 0, then all four possibilities clearly give nonnegative values for g. (For k = 0, we generate the first four values of your sequence: 0, 1, 6, 11.) Suppose we want to generate a list of such values: We can exhaust all the possibilities for g by evaluating the above four formulas for g for each k = 0,1,2,3,... (When writing your list, because you need positive integer values for g, you'll need to throw away the first element of your list, 12(0)^2+(0)=0.) Considering the values of g generated by a particular k, we have (clearly): 12k^2+k < 12k^2+7k+1 < 12k^2+17k+6 < 12k^2+23k+11. So, the smallest value of g generated by k+1 is: 12(k+1)^2+(k+1) = 12k^2+25k+13 > 12k^2+23k+11 That is, the smallest value for g generated by k+1 is larger than the largest value generated by k. So, we can write down a complete ordered list (a sequence that gives the possible solutions for g in increasing order) by writing down the four possibilities for g generated by 0 (in increasing order), then the four generated by 1, then those for 2, and so on. I don't know how to write the sequence more compactly than by writing down the above rule. Travis > Hi all, > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? > cdj === Subject: Integral Hi. What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral >Hi. >What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? http://integrals.wolfram.com/ says: RootSum[-1+#1+#1^2+#1^5 &, (Log[x-#1] / (1+2#1+5#1^4)) &] Thomas === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? As soon as you come up with a closed form of factorization of x^5 + x^2 + x - 1 (one linear factor and two quadratic factors), I will be able to tell you more. === Subject: Re: Integral Mike4ty, Ugly, I think. Factor the polynomial (warning, it's irreducible over the integers), then apply the method of partial fractions. Travis > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? The hard part is solving the quintic. There is a real root near 0.568544, a complex conjugate pair near 0.91612 +/- 0.57771 i, and another pair near 0.622848 +/- 1.03222 i. Once you have the factors, it's an easy partial fractions decomposition, provided you don't mind approximate answers. Mathematica 5.0 for Mac OS X -- Terminal graphics initialized -- In[1]:= Integrate[1./(x^5+x^2+x-1),x] Out[1]= 1. (-0.22894 ArcTan[0.484391 (-1.2457 + 2. x)] - > 0.189874 ArcTan[0.865487 (1.83224 + 2. x)] + > 0.361679 Log[0.586544 - 1. x] - 2 > 0.0641575 Log[1.45342 - 1.2457 x + x ] - 2 > 0.116682 Log[1.17302 + 1.83224 x + x ]) -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Galois group = A_4 The following question has been bugging me : I am trying to come up with a quartic polynomial whose Galois group is A_4. I know I can go about the business of finding a discriminant that is the square of a rational number, and the resolvent cubic is irreducible. But I was wondering if there was a more illuminating way to geometrically come up with a quartic polynomial. In particular, we know A_4 has no transpositions and no 4-cycles. So, what can I say about the quartic? Wouldn't it be true that since there are no transpositions, the quartic cannot have exactly 2 real roots? What can I deduce about the fact that the Galois group has no 4-cycles? Tony === Subject: Re: How to find the extremum of the Absolute value of a function=?big5?Q?=EF=BC=9F?= My method is to differentiate |Z| w.r.t y, and then subtitute y1 and y2 into the above equation. Finally I got a simultaneous equations and solving for x1 and x2. But where confused me is that the derivative of an absolute value seems not exist, so my known method didn't work due to the absolute value. is there any other method to solve a extremum problem of an obsolute value? Á¡ (Randy Poe)ÁnÛ¤Èæ2 50´ÁG > Suppose I have a complex-valued function Z, and Z=3DZ(x1,x2,y) > where x1,x2 and y are three real variables. I wanna |Z| has local > minimums at two given points y=3Dy1 and y=3Dy2, where x1 and x2 should be > adjusted to met this demand. > i.e. > Q=EF=BC=9AHow to find x1 and x2 such that |Z| has local minimum at two > given points > y1 and y2 ? > To find the minimimum for y=3Dy1, define a new function: > W1(x1,x2) =3D Z(x1, x2, y1) > and use your favorite minimization technique. Similarly for y2. > - Randy -- Á¡ Origin: ´.beÛjË.b3.bc®[UDo ubleDot].90[Degre e]Tøü  === Subject: swjpam fyi, The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer exists due to budgetary contraints. === Subject: Re: swjpam Discussion, linux) > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. I will never doubt the hammer again. Golly. -- Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others. -- Ambrose Bierce === Subject: Re: swjpam > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. Is that just the excuse? Is the real reason editorial incompetence? Is this the first blow of The Hammer? === Subject: do you have any smart way of finding which number is bigger ? Using the fatest way: compare: 0.9^10 vs. 2*(0.9^19)+0.9^20 how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? lucy escribi.97: > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? 2*(0.9^19)+0.9^20 = 0.9^10(2*0.9^9 + 0.9^10) = = 0.9^10(20*9^9 + 9^10)/10^10 = 0.9^10* 9^9*29/10^10 But 9^2 = 81 > 80, then 9^8 > 8^4*10^4 = 4096*10^4 ==> 9^9 > 36*10^7 ==> 29*9^9 > 36*29*10^7 = 1044*10^7 > 10^10 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to over 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? A: 0.9^10 B: 2*(0.9^19)+0.9^20 Notice both pieces of B are positive. b1 > 0 b2>0 So if I can see one side of + larger than A then B > A 0.9^20 > 0.9^10 20 > 10 ergo B > A 5 seconds === Subject: Re: do you have any smart way of finding which number is bigger ? lucy, Here's a way that doesn't use much explicit calculation. (It admittedly uses the fact that e < 2.9, which can easily be derived analytically, anyway). The Taylor series for log about 1 is: log(1+x) = x - (1/2)x^2 + (1/3)x^3 - ... Setting x = 1/9, we have: log(10/9) = log(1 + 1/9) = (1/9) - (1/2)(1/9)^2 + (1/3)(1/9)^3 - ... < 1/9 9 * log (10/9) < 1 9 log 10 - 9 log 9 < 1 Adding log 10, 10 log 10 - 9 log 9 < 1 + log 10 = log (e * 10) < log (29) [Here is where I invoke e < 2.9.] Subtracting log 9, 10 log 10 - 10 log 9 < log 29 - log 9 10 log (10/9) < log (29/9) = log (2 * (10/9) + 1) Exponentiating, (10/9)^10 < 2*(10/9) + 1 Multiplying both sides by (9/10)^20 (9/10)^10 < 2*(9/10)^19 + (9/10)^20 I suspect there is a much more elegant way. Maybe something with the expression 10 log 10 - 9 log 9 < log (29)? Travis > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? I don't know if it's smart or fastest, but you can factor. 0.9^10 >?< 2 * (0.9^19) + 0.9^20 0.9^10 >?< 2 * (0.9^20)/0.9 + 0.9^20 0.9^10 >?< 0.9^20 * (2/0.9 + 1) 0.9^10 / 0.9^20 >?< 2/0.9 + 1 0.9^-10 >?< 2.2222... + 1 2.8679... < 3.2222... -- john === Subject: Re: do you have any smart way of finding which number is bigger ? ETAtAhUAvgUxhHwS5c+oabk25UmVuHI06JUCFAza+gLCc85dGRk3KQeD5aQy6jyx 0.9^10 ? 2*(0.9^19)+0.9^20 1 ? 2^(0.9^9) + 0.9^10 10^10 ? 20*9^9 + 9^10 = 29*9^9 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. Therefore ? is <. --OL === Subject: Re: do you have any smart way of finding which number is bigger ? > 0.9^10 ? 2*(0.9^19)+0.9^20 > 1 ? 2^(0.9^9) + 0.9^10 > 10^10 ? 20*9^9 + 9^10 = 29*9^9 I followed you up to this point; you are now comparing 10^10 with 29 * 9^9 I don't understand the next line. > 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. > Therefore ? is <. -- john === Subject: Re: do you have any smart way of finding which number is bigger ? .9^10 = 0.3486784401 2*(0.9^19) + 0.9^20 = 0.39174699812516770581 The second one is larger time to do the problem -- (2 seconds for cut and paste maybe?) === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I don't see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you don't believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you don't believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). --Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! you're not your Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didn't know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didn't know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Here's a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is r's n'th digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but that's not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker What's a hyper-real number? Do you even know anything about math? Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> What's a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least it's a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that I'm all that neat, but hopefully my notation's clear at least.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You don't even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You don't even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, don't look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. I'll define it with an example. Suppose you have alot of people here making noise here on this NG and they don't know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You don't even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, don't look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > I'll define it with an example. Suppose you have alot of people here >making >noise here on this NG and they don't know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry don't look... Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still don't know what a hyper-real number is. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: terms in Math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmj09356; >Can anyone tell me the URL with pages devoted to math term's. One such is http://mathworld.wolfram.com/ === Subject: Re: coverings of a Mobius band by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmT09372; >Are the even degree covers of a Mobius band annulus(es)? Are the odd degree covers of a Mobius band, Mobius bands again? What is the universal cover of a Mobius band? Yes, yes, and R x [-1, 1]. One can imagine a strip of paper with n half-twists before gluing the ends as the total space of the degree n cover of the Mobius band. The strip of paper is I x [-1, 1], and the gluing map for taping the ends is the homeomorphism (-1)^n * - : [-1, 1] --> [-1, 1], so we get an annulus for n even and a Mobius strip for n odd. For the universal cover, imagine an infinite strip with infinitely many half-twists. Todd Trimble === Subject: Re: The Possible Cure For AIDS in part: > There are no A's, G's, 4's, O's, >>The G is gimmel the 3 rd letter of the Hebrew alphabet. >>The vowels (mplicit in old Hebrew but made explicit by the Masorites) >>include an ah sound and a oh sound. >AFAIK the Torah codes are never done with the vowels. That's quite true. Including vowels makes it harder to find matches. But if someone wanted to find English words in the Tanakh, the standard convention is, of course, Aleph for A and Ayin for O. As for the number 404, two letters would represent that in standard Hebrew numeric notation: Tau and Daleth. But Hebrew has a word for silver, although *ancient* Hebrew hardly had a word for oxygen, even if it certainly has one today. Actually, we should be asking what letters of the Hebrew alphabet are supposed to stand for and ! John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Possible Cure For AIDS > But Hebrew has a word for silver, although *ancient* Hebrew hardly had a Kessef. > word for oxygen, even if it certainly has one today. Khamtzan which means sour-er. This corresponds to the German Saurstuff. > Actually, we should be asking what letters of the Hebrew alphabet are > supposed to stand for and ! There is a typefont used to crossreference passages in the Talmud to the Shulkhan Arooch. It looks like superscripts because of the way it is printed. This was originated about 1500 of the common era. Bob Kolker === Subject: Re: The Possible Cure For AIDS in part: >There are no A's, G's, 4's, O's, etc. in Hebrew. Did you transliterate >these according to some arbitrary scheme, or did you search for Hebrew >letter strings? Not that it matters, of course. But while we're talking about a cure for AIDS... Some years back, there were news stories about how, for AIDS research, modified mice were created with human immune systems. This was done by taking cells from human fetuses. Naturally, this was controversial because of the abortion issue. I noted that nobody cares about mouse fetuses. If you can fix a mouse so that it _can_ get AIDS by giving it a human being's immune system, it would seem you could fix a human so that he can't get AIDS by giving him a mouse's immune system. There were probably very good reasons why this couldn't _really_ be done, but I had thought it worth mentioning. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Beal's conjecture > => A large prize is offered by banker Beal for a solution to > => the Beal Conjecture: the equation x^p + y^q = z^r has no solutions > => for p, q, r > 2 and coprime integers x, y, z. > = => Sorry if this has been discussed here - my only justification > => is that I just recently discovered this group but I am curious to > => know if Wile's proof of FLT also covers the Beal Conjecture > => Or have any counter-examples been found? Wile's proof covered the case p=q=r and relaxed the coprime requirement for x, y, & z. IE: There is no integer solution when p=q=r whatever the coprimality of x, y, & z. tom -- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an hour. === Subject: Mathforge.net :: Near-numbers, autonomic computing, John Derbyshire, and Church-Turing The Latest Math News from Mathforge.net http://mathforge.net ****An invitation to additive prime number theory**** A.V. Kumchev and D.I. Tolev have compiled a short document entitled An Invitation to Additive Prime Number Theory[~60pp, pdf]. The document serves as an introductory guide to graduate-level students on... http://mathforge.net/index.jsp?page=seeReplies&messageNum=981 ****Are there encoded messages in the Bible?**** researchers both supporting and denying the statistical evidence for 'hidden messages' found in the Old Testament tell their tales. Some... http://mathforge.net/index.jsp?page=seeReplies&messageNum=974 ****Near-numbers: the new 'limit'**** There's a very interesting paper by Frank J Swenton of Middlebury College called Limits and the System of Near-Numbers[19pp, pdf]. What seemed at first glance (at the title and abstract) like an uninspired... http://mathforge.net/index.jsp?page=seeReplies&messageNum=970 ****Introduction to autonomic computing**** While not itself mathematical in nature, the concept is built around ideas garnered from areas of artificial intelligence research and it is easy to see that autonomic computing could have applications... http://mathforge.net/index.jsp?page=seeReplies&messageNum=969 ****Derbyshire's Diary**** John Derbyshire, author of Prime Obsession, has a mathematical problem accompanying each of his Diary entries located in his Web Journalism folder. If you sift through enough of the partisan propaganda... http://mathforge.net/index.jsp?page=seeReplies&messageNum=968 ****Maple 9.5 Released**** MapleSoft has released version 9.5 of their popular symbolic and numeric computational software suite Maple. New Features include added packages (optimization, logic, and root finding), OpenMaple access... http://mathforge.net/index.jsp?page=seeReplies&messageNum=967 ****Mathematica 5.1 released**** The Mathematica version has jumped a tenth of a point, and Wolfram has added scores of new features to the new release, including Web Services support, a benchmarking package, string manipulation functions,... http://mathforge.net/index.jsp?page=seeReplies&messageNum=966 ****Quantum computers and the Church-Turing Thesis**** The original Church-Turing Thesis states that every function which would naturally be regarded as computable can be computed by a Turing Machine and Petrus H. Potgeiter mentions in his paper Zeno Machines... http://mathforge.net/index.jsp?page=seeReplies&messageNum=963 ****Sobering U.S. Student Math Scores**** In a disheartening follow-up to the Putnam story below, news (Seattle Times) outlets (New York Times)everywhere are reporting the horrid state of U.S. student math skills. The Organisation for Economic... http://mathforge.net/index.jsp?page=seeReplies&messageNum=959 ''Mathforge ran a story about the Putnam Competition... http://mathforge.net/index.jsp?page=seeReplies&messageNum=958 === Subject: What kind of matrix can map positive element vectors to positive element vectors? Hi all, Suppose I have a matrix A, and a positive element vector x, then y=A*x, I want y to be also positive element vector... What can I say about A? I want all such kinds of A? What kinds of A can let me have both directions: x positive elements <=> y=A*x positive elements? === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > Suppose I have a matrix A, and a positive element vector x, > then y=A*x, > I want y to be also positive element vector... > What can I say about A? I want all such kinds of A? > What kinds of A can let me have both directions: > x positive elements <=> y=A*x positive elements? Take a simple case and try to learn from it. The unit vectors for x will pull out the columns of A. Sure seems to say that A must at least have all of its elements positive. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > > Suppose I have a matrix A, and a positive element vector x, > > then y=A*x, > > I want y to be also positive element vector... > > What can I say about A? I want all such kinds of A? > > What kinds of A can let me have both directions: > > x positive elements <=> y=A*x positive elements? > > > > Take a simple case and try to learn from it. > The unit vectors for x will pull out the columns of A. > Sure seems to say that A must at least have all of its > elements positive. Necessary and sufficient. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? >> Hi all, >> Suppose I have a matrix A, and a positive element vector x, >> then y=A*x, >> I want y to be also positive element vector... >> What can I say about A? I want all such kinds of A? >> What kinds of A can let me have both directions: >> x positive elements <=> y=A*x positive elements? >> Take a simple case and try to learn from it. >> The unit vectors for x will pull out the columns of A. >> Sure seems to say that A must at least have all of its >> elements positive. >Necessary and sufficient. Neither necessary (unless your positive means >= 0) nor sufficient. Note that lucy wanted <=>. I'll assume A is a square matrix. It's easy to see A must be nonsingular, else given vector x > 0 (i.e. all x_i > 0) with Ax > 0, you could add to x a suitable multiple of a vector w with Aw = 0 to get a vector v not > 0 with Av = Ax > 0. A must map the nonnegative cone C = {x in R^n: all x_i >= 0} onto itself. Note that the extreme rays of C pass through the standard unit vectors e(j) (with e(j)_i = 1 if i=j, 0 otherwise). That is, these are the only members w of C such that if w = t x + (1-t) y with x,y in C and 0 < t < 1, then x and y are scalar multiples of w. Now A must map extreme rays to extreme rays, and from this it's easy to see that A must be of the form A = D P where P is a permutation matrix and D a diagonal matrix with positive elements on the diagonal. Conversely, everything of this form has the desired property. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: How to determine parameter integer values such that quadratic has integer solutions? Hi all, In particular, I'm interested in finding the positive integer values of g such that the following quadratic in v has integer roots: v^2 - 7v + (12 - 12g) Evidently this requires that 1 + 48g be a perfect square. I can have Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, 20, 35, ... This sequence is more conveniently summarized as something-mod-something-else, but I forget exactly what the two somethings are. How to figure this out? cdj === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) v = (7 +- sqr(49 - 48 + 48g)/2 = (7 +- sqr(1 + 48g)/2 sqr(1 + 48g) must be odd number, 2n + 1 1 + 48g = 4n^2 + 4n + 1 12g = n^2 + n = n(n + 1) Case 12 | n. Done Case 4 | n, not 3 | n. 3 | n+1; n = 3k - 1 Case 3 | n, not 2 | n. 4 | n+1; n = 4k - 1 Case 2 | n, not 4 | n. Not possible Three family parametrization of solutions. n = 12k n = 3k - 1 provided 4 | n n = 4k - 1 provided 3 | n Expect some duplicates. > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? The square root of 1+48g is odd... for fuck's sake I'm blind.... === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? cdj: Here's an outline: We know that 1+48g is a perfect square. Since g is an integer, 48g is even, and 1+48g is odd. So, the square root of 1+48g is odd, and we can write that square root as 2n+1 for some integer n. Thus, (2n+1)^2 = 1+48g Expanding and simplifying, 4n^2+4n+1 = 1+48g 4n^2+4n = 48g n^2+n = 12g We need a little number theory here: Consider the above equation (mod 12): (n^2+n) (mod 12) = 0 Trying the 11 possibilities, we find that n (mod 12) must be 0, 3, 8, or 11, that is, we must be able to write n as 12k or 12k+3 or 12k+8 or 12k+11 for some integer k. (The above condition is equivalent to (n^2+n) = 0 (mod 3) AND (n^2+n) = 0 (mod 4), so with some work, we really need only to consider 3+4 = 7 cases.) Substituting for the first possibility (n = 12k, that is, n is a multiple of 12), n^2+n = (12k)^2+(12k) = 144k^2+12k = 12(12k^2+k) But, n^2+n = 12g, so this case gives: 12(k^2+k) = 12g, that is, g = 12k^2+k The second possibility (n = 12k+3, that is, n is 3 more than a multiple of 12) similarly gives n^2+n = (12k+3)^2+(12k+3) = 144k^2+84k+12 = 12(12k^2+7k+1) 12(12k^2+7k+1) = 12g, so, g = 12k^2+7k+1 The other possibilities (n = 12k+8, n=12k+11) give after similar calculations, g = 12k^2+17k+6, and g = 12k^2+23k+11 If k > 0, then all four possibilities clearly give nonnegative values for g. (For k = 0, we generate the first four values of your sequence: 0, 1, 6, 11.) Suppose we want to generate a list of such values: We can exhaust all the possibilities for g by evaluating the above four formulas for g for each k = 0,1,2,3,... (When writing your list, because you need positive integer values for g, you'll need to throw away the first element of your list, 12(0)^2+(0)=0.) Considering the values of g generated by a particular k, we have (clearly): 12k^2+k < 12k^2+7k+1 < 12k^2+17k+6 < 12k^2+23k+11. So, the smallest value of g generated by k+1 is: 12(k+1)^2+(k+1) = 12k^2+25k+13 > 12k^2+23k+11 That is, the smallest value for g generated by k+1 is larger than the largest value generated by k. So, we can write down a complete ordered list (a sequence that gives the possible solutions for g in increasing order) by writing down the four possibilities for g generated by 0 (in increasing order), then the four generated by 1, then those for 2, and so on. I don't know how to write the sequence more compactly than by writing down the above rule. Travis > Hi all, > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? > cdj === Subject: Integral Hi. What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral >Hi. >What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? http://integrals.wolfram.com/ says: RootSum[-1+#1+#1^2+#1^5 &, (Log[x-#1] / (1+2#1+5#1^4)) &] Thomas === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? As soon as you come up with a closed form of factorization of x^5 + x^2 + x - 1 (one linear factor and two quadratic factors), I will be able to tell you more. === Subject: Re: Integral Mike4ty, Ugly, I think. Factor the polynomial (warning, it's irreducible over the integers), then apply the method of partial fractions. Travis > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? The hard part is solving the quintic. There is a real root near 0.568544, a complex conjugate pair near 0.91612 +/- 0.57771 i, and another pair near 0.622848 +/- 1.03222 i. Once you have the factors, it's an easy partial fractions decomposition, provided you don't mind approximate answers. Mathematica 5.0 for Mac OS X -- Terminal graphics initialized -- In[1]:= Integrate[1./(x^5+x^2+x-1),x] Out[1]= 1. (-0.22894 ArcTan[0.484391 (-1.2457 + 2. x)] - > 0.189874 ArcTan[0.865487 (1.83224 + 2. x)] + > 0.361679 Log[0.586544 - 1. x] - 2 > 0.0641575 Log[1.45342 - 1.2457 x + x ] - 2 > 0.116682 Log[1.17302 + 1.83224 x + x ]) -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Galois group = A_4 The following question has been bugging me : I am trying to come up with a quartic polynomial whose Galois group is A_4. I know I can go about the business of finding a discriminant that is the square of a rational number, and the resolvent cubic is irreducible. But I was wondering if there was a more illuminating way to geometrically come up with a quartic polynomial. In particular, we know A_4 has no transpositions and no 4-cycles. So, what can I say about the quartic? Wouldn't it be true that since there are no transpositions, the quartic cannot have exactly 2 real roots? What can I deduce about the fact that the Galois group has no 4-cycles? Tony === Subject: Re: How to find the extremum of the Absolute value of a function=?big5?Q?=EF=BC=9F?= My method is to differentiate |Z| w.r.t y, and then subtitute y1 and y2 into the above equation. Finally I got a simultaneous equations and solving for x1 and x2. But where confused me is that the derivative of an absolute value seems not exist, so my known method didn't work due to the absolute value. is there any other method to solve a extremum problem of an obsolute value? Á¡ (Randy Poe)ÁnÛ¤Èæ2 50´ÁG > Suppose I have a complex-valued function Z, and Z=3DZ(x1,x2,y) > where x1,x2 and y are three real variables. I wanna |Z| has local > minimums at two given points y=3Dy1 and y=3Dy2, where x1 and x2 should be > adjusted to met this demand. > i.e. > Q=EF=BC=9AHow to find x1 and x2 such that |Z| has local minimum at two > given points > y1 and y2 ? > To find the minimimum for y=3Dy1, define a new function: > W1(x1,x2) =3D Z(x1, x2, y1) > and use your favorite minimization technique. Similarly for y2. > - Randy -- Á¡ Origin: ´.beÛjË.b3.bc®[UDo ubleDot].90[Degre e]Tøü  === Subject: swjpam fyi, The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer exists due to budgetary contraints. === Subject: Re: swjpam Discussion, linux) > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. I will never doubt the hammer again. Golly. -- Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others. -- Ambrose Bierce === Subject: Re: swjpam > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. Is that just the excuse? Is the real reason editorial incompetence? Is this the first blow of The Hammer? === Subject: do you have any smart way of finding which number is bigger ? Using the fatest way: compare: 0.9^10 vs. 2*(0.9^19)+0.9^20 how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? lucy escribi.97: > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? 2*(0.9^19)+0.9^20 = 0.9^10(2*0.9^9 + 0.9^10) = = 0.9^10(20*9^9 + 9^10)/10^10 = 0.9^10* 9^9*29/10^10 But 9^2 = 81 > 80, then 9^8 > 8^4*10^4 = 4096*10^4 ==> 9^9 > 36*10^7 ==> 29*9^9 > 36*29*10^7 = 1044*10^7 > 10^10 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to over 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? A: 0.9^10 B: 2*(0.9^19)+0.9^20 Notice both pieces of B are positive. b1 > 0 b2>0 So if I can see one side of + larger than A then B > A 0.9^20 > 0.9^10 20 > 10 ergo B > A 5 seconds === Subject: Re: do you have any smart way of finding which number is bigger ? lucy, Here's a way that doesn't use much explicit calculation. (It admittedly uses the fact that e < 2.9, which can easily be derived analytically, anyway). The Taylor series for log about 1 is: log(1+x) = x - (1/2)x^2 + (1/3)x^3 - ... Setting x = 1/9, we have: log(10/9) = log(1 + 1/9) = (1/9) - (1/2)(1/9)^2 + (1/3)(1/9)^3 - ... < 1/9 9 * log (10/9) < 1 9 log 10 - 9 log 9 < 1 Adding log 10, 10 log 10 - 9 log 9 < 1 + log 10 = log (e * 10) < log (29) [Here is where I invoke e < 2.9.] Subtracting log 9, 10 log 10 - 10 log 9 < log 29 - log 9 10 log (10/9) < log (29/9) = log (2 * (10/9) + 1) Exponentiating, (10/9)^10 < 2*(10/9) + 1 Multiplying both sides by (9/10)^20 (9/10)^10 < 2*(9/10)^19 + (9/10)^20 I suspect there is a much more elegant way. Maybe something with the expression 10 log 10 - 9 log 9 < log (29)? Travis > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? I don't know if it's smart or fastest, but you can factor. 0.9^10 >?< 2 * (0.9^19) + 0.9^20 0.9^10 >?< 2 * (0.9^20)/0.9 + 0.9^20 0.9^10 >?< 0.9^20 * (2/0.9 + 1) 0.9^10 / 0.9^20 >?< 2/0.9 + 1 0.9^-10 >?< 2.2222... + 1 2.8679... < 3.2222... -- john === Subject: Re: do you have any smart way of finding which number is bigger ? ETAtAhUAvgUxhHwS5c+oabk25UmVuHI06JUCFAza+gLCc85dGRk3KQeD5aQy6jyx 0.9^10 ? 2*(0.9^19)+0.9^20 1 ? 2^(0.9^9) + 0.9^10 10^10 ? 20*9^9 + 9^10 = 29*9^9 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. Therefore ? is <. --OL === Subject: Re: do you have any smart way of finding which number is bigger ? > 0.9^10 ? 2*(0.9^19)+0.9^20 > 1 ? 2^(0.9^9) + 0.9^10 > 10^10 ? 20*9^9 + 9^10 = 29*9^9 I followed you up to this point; you are now comparing 10^10 with 29 * 9^9 I don't understand the next line. > 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. > Therefore ? is <. -- john === Subject: Re: do you have any smart way of finding which number is bigger ? .9^10 = 0.3486784401 2*(0.9^19) + 0.9^20 = 0.39174699812516770581 The second one is larger time to do the problem -- (2 seconds for cut and paste maybe?) === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I don't see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you don't believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you don't believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smart's Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). --Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you don't believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = floor(x) + sum((floor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smart's Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! you're not your Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didn't know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didn't know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Here's a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is r's n'th digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but that's not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker What's a hyper-real number? Do you even know anything about math? Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> What's a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least it's a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that I'm all that neat, but hopefully my notation's clear at least.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You don't even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You don't even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, don't look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. I'll define it with an example. Suppose you have alot of people here making noise here on this NG and they don't know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > What's a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You don't even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, don't look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > I'll define it with an example. Suppose you have alot of people here >making >noise here on this NG and they don't know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry don't look... Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> What's a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You don't even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, don't look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still don't know what a hyper-real number is. >Bob Kolker Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas === Subject: Re: terms in Math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmj09356; >Can anyone tell me the URL with pages devoted to math term's. One such is http://mathworld.wolfram.com/ === Subject: Re: coverings of a Mobius band by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmT09372; >Are the even degree covers of a Mobius band annulus(es)? Are the odd degree covers of a Mobius band, Mobius bands again? What is the universal cover of a Mobius band? Yes, yes, and R x [-1, 1]. One can imagine a strip of paper with n half-twists before gluing the ends as the total space of the degree n cover of the Mobius band. The strip of paper is I x [-1, 1], and the gluing map for taping the ends is the homeomorphism (-1)^n * - : [-1, 1] --> [-1, 1], so we get an annulus for n even and a Mobius strip for n odd. For the universal cover, imagine an infinite strip with infinitely many half-twists. Todd Trimble === Subject: Re: The Possible Cure For AIDS in part: > There are no A's, G's, 4's, O's, >>The G is gimmel the 3 rd letter of the Hebrew alphabet. >>The vowels (mplicit in old Hebrew but made explicit by the Masorites) >>include an ah sound and a oh sound. >AFAIK the Torah codes are never done with the vowels. That's quite true. Including vowels makes it harder to find matches. But if someone wanted to find English words in the Tanakh, the standard convention is, of course, Aleph for A and Ayin for O. As for the number 404, two letters would represent that in standard Hebrew numeric notation: Tau and Daleth. But Hebrew has a word for silver, although *ancient* Hebrew hardly had a word for oxygen, even if it certainly has one today. Actually, we should be asking what letters of the Hebrew alphabet are supposed to stand for and ! John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Possible Cure For AIDS > But Hebrew has a word for silver, although *ancient* Hebrew hardly had a Kessef. > word for oxygen, even if it certainly has one today. Khamtzan which means sour-er. This corresponds to the German Saurstuff. > Actually, we should be asking what letters of the Hebrew alphabet are > supposed to stand for and ! There is a typefont used to crossreference passages in the Talmud to the Shulkhan Arooch. It looks like superscripts because of the way it is printed. This was originated about 1500 of the common era. Bob Kolker === Subject: Re: The Possible Cure For AIDS in part: >There are no A's, G's, 4's, O's, etc. in Hebrew. Did you transliterate >these according to some arbitrary scheme, or did you search for Hebrew >letter strings? Not that it matters, of course. But while we're talking about a cure for AIDS... Some years back, there were news stories about how, for AIDS research, modified mice were created with human immune systems. This was done by taking cells from human fetuses. Naturally, this was controversial because of the abortion issue. I noted that nobody cares about mouse fetuses. If you can fix a mouse so that it _can_ get AIDS by giving it a human being's immune system, it would seem you could fix a human so that he can't get AIDS by giving him a mouse's immune system. There were probably very good reasons why this couldn't _really_ be done, but I had thought it worth mentioning. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Beal's conjecture > => A large prize is offered by banker Beal for a solution to > => the Beal Conjecture: the equation x^p + y^q = z^r has no solutions > => for p, q, r > 2 and coprime integers x, y, z. > = => Sorry if this has been discussed here - my only justification > => is that I just recently discovered this group but I am curious to > => know if Wile's proof of FLT also covers the Beal Conjecture > => Or have any counter-examples been found? Wile's proof covered the case p=q=r and relaxed the coprime requirement for x, y, & z. IE: There is no integer solution when p=q=r whatever the coprimality of x, y, & z. tom -- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an hour. === Subject: Mathforge.net :: Near-numbers, autonomic computing, John Derbyshire, and Church-Turing The Latest Math News from Mathforge.net http://mathforge.net ****An invitation to additive prime number theory**** A.V. Kumchev and D.I. Tolev have compiled a short document entitled An Invitation to Additive Prime Number Theory[~60pp, pdf]. The document serves as an introductory guide to graduate-level students on... http://mathforge.net/index.jsp?page=seeReplies&messageNum=981 ****Are there encoded messages in the Bible?**** researchers both supporting and denying the statistical evidence for 'hidden messages' found in the Old Testament tell their tales. Some... http://mathforge.net/index.jsp?page=seeReplies&messageNum=974 ****Near-numbers: the new 'limit'**** There's a very interesting paper by Frank J Swenton of Middlebury College called Limits and the System of Near-Numbers[19pp, pdf]. What seemed at first glance (at the title and abstract) like an uninspired... http://mathforge.net/index.jsp?page=seeReplies&messageNum=970 ****Introduction to autonomic computing**** While not itself mathematical in nature, the concept is built around ideas garnered from areas of artificial intelligence research and it is easy to see that autonomic computing could have applications... http://mathforge.net/index.jsp?page=seeReplies&messageNum=969 ****Derbyshire's Diary**** John Derbyshire, author of Prime Obsession, has a mathematical problem accompanying each of his Diary entries located in his Web Journalism folder. If you sift through enough of the partisan propaganda... http://mathforge.net/index.jsp?page=seeReplies&messageNum=968 ****Maple 9.5 Released**** MapleSoft has released version 9.5 of their popular symbolic and numeric computational software suite Maple. New Features include added packages (optimization, logic, and root finding), OpenMaple access... http://mathforge.net/index.jsp?page=seeReplies&messageNum=967 ****Mathematica 5.1 released**** The Mathematica version has jumped a tenth of a point, and Wolfram has added scores of new features to the new release, including Web Services support, a benchmarking package, string manipulation functions,... http://mathforge.net/index.jsp?page=seeReplies&messageNum=966 ****Quantum computers and the Church-Turing Thesis**** The original Church-Turing Thesis states that every function which would naturally be regarded as computable can be computed by a Turing Machine and Petrus H. Potgeiter mentions in his paper Zeno Machines... http://mathforge.net/index.jsp?page=seeReplies&messageNum=963 ****Sobering U.S. Student Math Scores**** In a disheartening follow-up to the Putnam story below, news (Seattle Times) outlets (New York Times)everywhere are reporting the horrid state of U.S. student math skills. The Organisation for Economic... http://mathforge.net/index.jsp?page=seeReplies&messageNum=959 ''Mathforge ran a story about the Putnam Competition... http://mathforge.net/index.jsp?page=seeReplies&messageNum=958 === Subject: What kind of matrix can map positive element vectors to positive element vectors? Hi all, Suppose I have a matrix A, and a positive element vector x, then y=A*x, I want y to be also positive element vector... What can I say about A? I want all such kinds of A? What kinds of A can let me have both directions: x positive elements <=> y=A*x positive elements? === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > Suppose I have a matrix A, and a positive element vector x, > then y=A*x, > I want y to be also positive element vector... > What can I say about A? I want all such kinds of A? > What kinds of A can let me have both directions: > x positive elements <=> y=A*x positive elements? Take a simple case and try to learn from it. The unit vectors for x will pull out the columns of A. Sure seems to say that A must at least have all of its elements positive. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > > Suppose I have a matrix A, and a positive element vector x, > > then y=A*x, > > I want y to be also positive element vector... > > What can I say about A? I want all such kinds of A? > > What kinds of A can let me have both directions: > > x positive elements <=> y=A*x positive elements? > > > > Take a simple case and try to learn from it. > The unit vectors for x will pull out the columns of A. > Sure seems to say that A must at least have all of its > elements positive. Necessary and sufficient. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? >> Hi all, >> Suppose I have a matrix A, and a positive element vector x, >> then y=A*x, >> I want y to be also positive element vector... >> What can I say about A? I want all such kinds of A? >> What kinds of A can let me have both directions: >> x positive elements <=> y=A*x positive elements? >> Take a simple case and try to learn from it. >> The unit vectors for x will pull out the columns of A. >> Sure seems to say that A must at least have all of its >> elements positive. >Necessary and sufficient. Neither necessary (unless your positive means >= 0) nor sufficient. Note that lucy wanted <=>. I'll assume A is a square matrix. It's easy to see A must be nonsingular, else given vector x > 0 (i.e. all x_i > 0) with Ax > 0, you could add to x a suitable multiple of a vector w with Aw = 0 to get a vector v not > 0 with Av = Ax > 0. A must map the nonnegative cone C = {x in R^n: all x_i >= 0} onto itself. Note that the extreme rays of C pass through the standard unit vectors e(j) (with e(j)_i = 1 if i=j, 0 otherwise). That is, these are the only members w of C such that if w = t x + (1-t) y with x,y in C and 0 < t < 1, then x and y are scalar multiples of w. Now A must map extreme rays to extreme rays, and from this it's easy to see that A must be of the form A = D P where P is a permutation matrix and D a diagonal matrix with positive elements on the diagonal. Conversely, everything of this form has the desired property. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: How to determine parameter integer values such that quadratic has integer solutions? Hi all, In particular, I'm interested in finding the positive integer values of g such that the following quadratic in v has integer roots: v^2 - 7v + (12 - 12g) Evidently this requires that 1 + 48g be a perfect square. I can have Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, 20, 35, ... This sequence is more conveniently summarized as something-mod-something-else, but I forget exactly what the two somethings are. How to figure this out? cdj === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) v = (7 +- sqr(49 - 48 + 48g)/2 = (7 +- sqr(1 + 48g)/2 sqr(1 + 48g) must be odd number, 2n + 1 1 + 48g = 4n^2 + 4n + 1 12g = n^2 + n = n(n + 1) Case 12 | n. Done Case 4 | n, not 3 | n. 3 | n+1; n = 3k - 1 Case 3 | n, not 2 | n. 4 | n+1; n = 4k - 1 Case 2 | n, not 4 | n. Not possible Three family parametrization of solutions. n = 12k n = 3k - 1 provided 4 | n n = 4k - 1 provided 3 | n Expect some duplicates. > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? The square root of 1+48g is odd... for fuck's sake I'm blind.... === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? cdj: Here's an outline: We know that 1+48g is a perfect square. Since g is an integer, 48g is even, and 1+48g is odd. So, the square root of 1+48g is odd, and we can write that square root as 2n+1 for some integer n. Thus, (2n+1)^2 = 1+48g Expanding and simplifying, 4n^2+4n+1 = 1+48g 4n^2+4n = 48g n^2+n = 12g We need a little number theory here: Consider the above equation (mod 12): (n^2+n) (mod 12) = 0 Trying the 11 possibilities, we find that n (mod 12) must be 0, 3, 8, or 11, that is, we must be able to write n as 12k or 12k+3 or 12k+8 or 12k+11 for some integer k. (The above condition is equivalent to (n^2+n) = 0 (mod 3) AND (n^2+n) = 0 (mod 4), so with some work, we really need only to consider 3+4 = 7 cases.) Substituting for the first possibility (n = 12k, that is, n is a multiple of 12), n^2+n = (12k)^2+(12k) = 144k^2+12k = 12(12k^2+k) But, n^2+n = 12g, so this case gives: 12(k^2+k) = 12g, that is, g = 12k^2+k The second possibility (n = 12k+3, that is, n is 3 more than a multiple of 12) similarly gives n^2+n = (12k+3)^2+(12k+3) = 144k^2+84k+12 = 12(12k^2+7k+1) 12(12k^2+7k+1) = 12g, so, g = 12k^2+7k+1 The other possibilities (n = 12k+8, n=12k+11) give after similar calculations, g = 12k^2+17k+6, and g = 12k^2+23k+11 If k > 0, then all four possibilities clearly give nonnegative values for g. (For k = 0, we generate the first four values of your sequence: 0, 1, 6, 11.) Suppose we want to generate a list of such values: We can exhaust all the possibilities for g by evaluating the above four formulas for g for each k = 0,1,2,3,... (When writing your list, because you need positive integer values for g, you'll need to throw away the first element of your list, 12(0)^2+(0)=0.) Considering the values of g generated by a particular k, we have (clearly): 12k^2+k < 12k^2+7k+1 < 12k^2+17k+6 < 12k^2+23k+11. So, the smallest value of g generated by k+1 is: 12(k+1)^2+(k+1) = 12k^2+25k+13 > 12k^2+23k+11 That is, the smallest value for g generated by k+1 is larger than the largest value generated by k. So, we can write down a complete ordered list (a sequence that gives the possible solutions for g in increasing order) by writing down the four possibilities for g generated by 0 (in increasing order), then the four generated by 1, then those for 2, and so on. I don't know how to write the sequence more compactly than by writing down the above rule. Travis > Hi all, > In particular, I'm interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? > cdj === Subject: Integral Hi. What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral >Hi. >What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? http://integrals.wolfram.com/ says: RootSum[-1+#1+#1^2+#1^5 &, (Log[x-#1] / (1+2#1+5#1^4)) &] Thomas === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? As soon as you come up with a closed form of factorization of x^5 + x^2 + x - 1 (one linear factor and two quadratic factors), I will be able to tell you more. === Subject: Re: Integral Mike4ty, Ugly, I think. Factor the polynomial (warning, it's irreducible over the integers), then apply the method of partial fractions. Travis > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? The hard part is solving the quintic. There is a real root near 0.568544, a complex conjugate pair near 0.91612 +/- 0.57771 i, and another pair near 0.622848 +/- 1.03222 i. Once you have the factors, it's an easy partial fractions decomposition, provided you don't mind approximate answers. Mathematica 5.0 for Mac OS X -- Terminal graphics initialized -- In[1]:= Integrate[1./(x^5+x^2+x-1),x] Out[1]= 1. (-0.22894 ArcTan[0.484391 (-1.2457 + 2. x)] - > 0.189874 ArcTan[0.865487 (1.83224 + 2. x)] + > 0.361679 Log[0.586544 - 1. x] - 2 > 0.0641575 Log[1.45342 - 1.2457 x + x ] - 2 > 0.116682 Log[1.17302 + 1.83224 x + x ]) -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Galois group = A_4 The following question has been bugging me : I am trying to come up with a quartic polynomial whose Galois group is A_4. I know I can go about the business of finding a discriminant that is the square of a rational number, and the resolvent cubic is irreducible. But I was wondering if there was a more illuminating way to geometrically come up with a quartic polynomial. In particular, we know A_4 has no transpositions and no 4-cycles. So, what can I say about the quartic? Wouldn't it be true that since there are no transpositions, the quartic cannot have exactly 2 real roots? What can I deduce about the fact that the Galois group has no 4-cycles? Tony === Subject: Re: How to find the extremum of the Absolute value of a function=?big5?Q?=EF=BC=9F?= My method is to differentiate |Z| w.r.t y, and then subtitute y1 and y2 into the above equation. Finally I got a simultaneous equations and solving for x1 and x2. But where confused me is that the derivative of an absolute value seems not exist, so my known method didn't work due to the absolute value. is there any other method to solve a extremum problem of an obsolute value? Á¡ (Randy Poe)ÁnÛ¤Èæ2 50´ÁG > Suppose I have a complex-valued function Z, and Z=3DZ(x1,x2,y) > where x1,x2 and y are three real variables. I wanna |Z| has local > minimums at two given points y=3Dy1 and y=3Dy2, where x1 and x2 should be > adjusted to met this demand. > i.e. > Q=EF=BC=9AHow to find x1 and x2 such that |Z| has local minimum at two > given points > y1 and y2 ? > To find the minimimum for y=3Dy1, define a new function: > W1(x1,x2) =3D Z(x1, x2, y1) > and use your favorite minimization technique. Similarly for y2. > - Randy -- Á¡ Origin: ´.beÛjË.b3.bc®[UDo ubleDot].90[Degre e]Tøü  === Subject: swjpam fyi, The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer exists due to budgetary contraints. === Subject: Re: swjpam Discussion, linux) > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. I will never doubt the hammer again. Golly. -- Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others. -- Ambrose Bierce === Subject: Re: swjpam > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. Is that just the excuse? Is the real reason editorial incompetence? Is this the first blow of The Hammer? === Subject: do you have any smart way of finding which number is bigger ? Using the fatest way: compare: 0.9^10 vs. 2*(0.9^19)+0.9^20 how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? lucy escribi.97: > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? 2*(0.9^19)+0.9^20 = 0.9^10(2*0.9^9 + 0.9^10) = = 0.9^10(20*9^9 + 9^10)/10^10 = 0.9^10* 9^9*29/10^10 But 9^2 = 81 > 80, then 9^8 > 8^4*10^4 = 4096*10^4 ==> 9^9 > 36*10^7 ==> 29*9^9 > 36*29*10^7 = 1044*10^7 > 10^10 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? Err, the other thing you need to notice along this line is the 2* and the addition adding up to over 5x Of course 0.9^20 < 0.9^10